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THE UNIVERSITY OF AKRONMathematics and Computer Science

Algebra Review in Ten LessonsDirectory

• Lesson 1: Setting Up the Environment• Lesson 2: Exponents & Radicals• Lesson 3: Basic Algebra, Part I• Lesson 4: Basic Algebra, Part II• Lesson 5: Expansion• Lesson 6: Dividing & Factoring Polynomials• Lesson 7: Solving Equations & Inequalities• Lesson 8: Cartesian Coordinate System & Functions• Lesson 9: Functions (cont.) & First Degree Curves• Lesson 10: Some Second Degree & Trig Curves

Copyright c©1995–2000 D. P. StoryLast Revision Date: 2/2/2000

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Lesson 1: Setting Up the Environment

Directory

• Table of Contents• Begin Lesson 1

An

Algebra Review

I amDPSN ⊆ Z ⊆ Q ⊆ R ⊆ C

a3a4 = a7 (ab)10 = a10b10

−(ab − (3ab − 4)) = 2ab − 4(ab)3(a−1 + b−1) = (ab)2(a + b)(a − b)3 = a3 − 3a2b + 3ab2 − b3

2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }

f(x) = mx + by = sinx

inTen Lessons



Lesson 1: Setting Up the Environment

Table of Contents1. Setting Up the Environment1.1. The Real Number System, and friends

• The Natural Numbers • The Integers • The Rational Num-bers • The Irrational Numbers • The Real Numbers

1.2. The Number Line, and relations• Less than and Greater than

1.3. Distance and Absolute Value• Absolute Value • Distance between two numbers • TheMidpoint between two Numbers

1.4. Interval Notation

1. Setting Up the EnvironmentIn this lesson, we review some very basic ideas and terminology ofthe so-called Real Number System. Do not take this first lessonlightly, your knowledge of the real number system and its propertiesis key to your understanding why certain algebraic manipulationsare permissible, and why others are not. (When you are manipulatingalgebraic quantities, you are, in fact, manipulating numbers.)

Terminology is important in all scientific, technical and professionalfields, and mathematics is no different, it has a lot of it. When youspeak or write, you use words; the words you use must be under-stood by the ones with whom you are trying to communicate. Forsomeone to understand your communication, the words must have auniversal meaning; therefore, it is necessary for you to use the correctterminology to be able to effectively and efficiently communicate withothers. Correct use of (mathematical) terminology is a sign that youunderstand what you are saying or writing.

Section 1: Setting Up the Environment

1.1. The Real Number System, and friends

A pedestrian description of the real number system is that it is the setof all decimal numbers. Here are a few examples of decimal numbers:

1.456445 � Positive Number with Finite Decimal Expansion

−3.54 � Negative Number with Finite Decimal Expansion

1.49773487234 . . . � Positive Number with Infinite Decimal Expansion

−3.4 × 109� Negative Number in Scientific Notation

5.65445E−5 � Positive Number in Scientific Notation

You have been working with these representations of numbers allyour life. You have worked long and hard to master the mechanics ofadding, subtracting, multiplying, and dividing these numbers. Morerecently, the calculator makes many of the operations just mentionedautomatic and routine, but the calculator does not diminish the needto still have a ‘pencil and paper’ understanding of these operations,nor does it diminish the need to be able to read, interpret and convertnumbers.


Decimal numbers are, in fact, the end product of a series of prelim-inary definitions. As important as decimal numbers are, let’s leavethem for now to discuss a series of definitions that lead ultimately tothe construction of the real number system.

• The Natural NumbersThe natural numbers are the numbers

1, 2, 3, 4, 5, 6, . . . , 100, . . . , 1000, . . .

and so on ad infinitum. Mathematicians put braces around these num-bers to create the set of all natural numbers:

N = { 1, 2, 3, 4, 5, 6, . . . , n, . . . }. (1)

The letter N is typically used to denote this set; i.e., N symbolicallyrepresents the set of all natural numbers.


• The IntegersThe integers consist of the natural numbers, the negative naturalnumbers and 0; or more succinctly:

. . . ,−4,−3,−2,−1, 0, 1, 2, 3, 4, . . .

The set of all integers is denoted by Z (for some reason).

Z = { . . . ,−4,−3,−2,−1, 0, 1, 2, 3, 4, . . . }. (2)

The set of integers Z is an infinite set as well.

Question. Which of the following is true?(a) N ⊆ Z (b) Z ⊆ N

Question. If p is an integer, what can be said about 2p?(a) 2p is an odd number (b) 2p is an even number

An integer is either even or odd. Even and odd integers are often dealtwith in mathematics. It is important to have a general representationof these types of numbers. In the previous Question, we observed thatif p ∈ Z, then 2p is an even integer. An odd integer can be thought


as any integer that follows an even integer. Since, in general, 2p is aneven integer, 2p + 1 must be an odd integer. Thus,

2p is even, for any p ∈ Z

2p + 1 is odd, for any p ∈ Z(3)

Side Bar. Do you know what the symbol p ∈ Z means? Yes or No.

Question. Is the product of two odd integers, an odd integer or aneven integer?(a) Odd (b) Even (c) Can’t tell

Mathematics has a lot of little facts like the one above. You too candiscover interesting little tidbits—all you need is an inquiring mind.

• The Rational NumbersThe next step in the process of constructing the real number system isto create the rational number system. A rational number is a number


of the form p/q, where p, q ∈ Z and q �= 0. The letter used to denotethe set of all rational numbers is Q:

Q ={

p

q| p, q ∈ Z, q �= 0

}(4)

You have been dealing with rational numbers for most of your rationallife. Here are a few examples,

12,1223

, −34,1234554321

, − 812

. (5)

• Question. What is ‘wrong’ with the last number in list (5)? Thinkabout your answer then click on the green dot (it’s a bullet actually).

Rational numbers are important in the world of computers becausethese are the only kind of numbers a computer knows! To your calcu-lator, all numbers are rational.

Quiz. Here are some statements about rational numbers. Think beforeyou respond. A passing grade is 100%.


1. What is the relationship between Q and Z?(a) Q ⊆ Z (b) Z ⊆ Q (c) n.o.t.

2. Is the product of two rational numbers again a rational number?(a) False (b) True

3. Is the sum of two rational numbers again a rational number?(a) False (b) True

End Quiz.

Rationals and Decimals. Each rational number has a decimal rep-resentation.

1. Some rational numbers have a finite decimal expansion:

34

= 0.75 − 46125

= −0.368272

= 13.5

2. Some rational numbers have an infinite repeating decimal ex-pansion:

23

= 0.666666666 . . .123999

= 0.123123123123 . . .


3. The decimal expansion of some rational numbers is not unique;i.e., some rational numbers have two decimal expansions. Forexample,

34

= 0.75 = 0.74999999999999999 . . .

Generally, any rational number that has a finite decimal expan-sion also has an infinite repeating decimal expansion:

32341.424552 = 32341.42455199999999999 . . .

0.5 = 0.49999999999999 . . .

−12.43 = −12.429999999999 . . .

Do you get the idea?

Exercise 1.1. Write the decimal number 423.43537 as an infiniterepeating decimal.


Yeah, but what about this exercise? It may be a tricky one!

Exercise 1.2. Write the decimal number 56.0 as an infinite repeatingdecimal.

Any decimal number with an infinite repeating decimal expansion isa rational number. Do you know how to compute this number? (Yesor No) There is a nice little trick.

Example 1.1. Consider the decimal number:

453.123123123123123 . . .

Convert this number to a rational number.

Now try one yourself.

Exercise 1.3. Convert the infinite repeating decimal number

3.45454545454545 . . .

to a rational number.


Here’s a slightly trickier one.

Exercise 1.4. Convert the decimal number

4.177777777777777 . . .

to a rational number.

• The Irrational NumbersAs the old phrase goes, “All other numbers that are not rational arecalled irrational numbers.” But are there “other numbers”? Yes, ofcourse. The numbers

√2,

√3, and π are irrational numbers.

Characteristic of irrational numbers is their decimal expansions areinfinite and nonrepeating. One famous example is π; its decimal ex-pansion, or at least the first 30 digits, is

π = 3.14159265358979323846264338328 . . .

• The Real Numbers


The set of all real numbers, denoted by R, consists of all the rationalnumbers and all the irrational numbers. In symbols,

R = { x | x is a rational or an irrational number } (6)

In the sections that follow, some of the basic properties of the realnumber system are brought out. These properties form the basis forthe construction of an algebraic system.

1.2. The Number Line, and relations

The real number system, R, can be visualized by an horizontal line,typically called a number line. Usually, the number line is labeled usinga symbol such as x, y, or z; in this case the number line is called thex-axis, the y-axis, or the z-axis, respectively. We’ll label our numberline the x-axis.

-3 -2 -1 0 1 2 3x

The x-axisNumbers to the right of zero (0) are called positive numbers. Numbersto the left of zero (0) are called negative numbers.


-3 -2 -1 0 1 2 3x

The x-axisThe numbers in blue are the positive real numbers and the numbersin red are the negative real numbers.

• Less than and Greater thanLet a and b be distinct real numbers. (Distinct means a �= b.) We saythat a is less than b provided a − b is a negative number; in this casewe write a < b:

a < b means a − b is negative.

We say that a is greater than b provided a − b is a positive number;in this case we write a > b:

a > b means a − b is positive.

Variations on these definitions are a ≤ b (a is less than or equal to b)and a ≥ b (a is greater than or equal to b).


Geometrically, what do these relations mean? On the x-axis, a < b isinterpreted as the number a lies to the left of b.

a bx

Geometry of a < b

Similarly, the geometry of a > b is that a lies to the right of b.

b ax

Geometry of a > b

Of course, if a lies to the left of b (a < b) then b must lie to the rightof a (b > a)! Understand?

Here are a couple of questions to test your knowledge of inequalities.Read the questions carefully, draw an x-axis to illustrate the situation,and then respond to the question. Don’t guess. A passing score is100%.

Question. If the number c lies to the right of the number d, then(a) c < d (b) d < c (c) c ≤ d (d) d ≤ c


Question. If the number c does not lie to the left of the number d then(a) c < d (b) d < c (c) c ≤ d (d) d ≤ c

Hopefully the use of different letters did not disturb you psychologi-cally.

Double Sided Inequalities. A double inequality is an inequality ofthe form a ≤ x ≤ b, where a, b, and x are numbers. The meaning ofthis symbolism is

a ≤ x ≤ b means a ≤ x and x ≤ b,

which can be rephrased as

a ≤ x ≤ b means x is between a and b.

Exercise 1.5. Critique the following double inequality: 1 ≤ x ≤ −2.


1.3. Distance and Absolute Value

In mathematics, we often want to compute the distance one numberis away from another. As a first case, let’s compute the distance anygiven number is away from zero (0); this is a natural way to introducethe absolute value of a number.

• Absolute ValueHow do you calculate the distance one number is away from zero?Begin by example. The number 5 is ‘5 units’ away from 0. The number54.233 is ‘54.233 units’ units away from 0. That was simple. Nowfor negative numbers. The number −10 is ‘10 units’ from 0 and thenumber −43.6 is ‘43.6 units’ from 0.

There is a pattern here. If x is a positive number, then x is ‘x units’away from 0. Now if x is a negative number, the distance from 0 isobtained by ‘dropping’ the negative sign on the number x. The waywe can do this is by prefixing the number x with a negative sign. Thatnegative sign will cancel with the negative sign built into the numberx. (For example, the negative sign in −5 can be removed by prefixing a


negative sign: −(−5) = 5.) Returning to generality, a negative numberx is ‘−x units’ away from 0.

Summary. Summarizing the previous paragraph we have

The distance between x and 0 is ={

x if x ≥ 0−x if x < 0

The case study on the right-hand of the above equation is an impor-tant and deserves its own definition.

Absolute Value. Let x be a number, the absolute value of x,denoted by |x|, and is defined by

|x| ={

x if x ≥ 0−x if x < 0

(7)


This function is often a source of a lot of confusion for students. Youmust try to understand it on two levels: The Geometric interpretationof absolute value, and the technical definition (equation (7)).

When working with a numerical value, usually there is no problemwith using the absolute value: Who could miscalculate

|4.3| = 4.3 or | − 6.24| = 6.24

incorrectly? Problems are encountered when we are working with sym-bolic quantities.

Example 1.2. Suppose a is a number such that a > 5. Compute|a − 5| and |5 − a|.Then working with absolute values involving symbolic quantities, toremove the absolute values we must know the sign of this symbolicquantity. In Example 1.2, we could remove the absolute value from|a − 5| because we knew that a − 5 > 0.

Here’s an easy one for you.


Exercise 1.6. Suppose w < 4. Compute |w − 4|.A slight variation on the same theme is demonstrated in the next . . .

Example 1.3. Suppose s is a number such that s > 6. Compute|2s − 3|.Now you investigate the following question.

Exercise 1.7. Suppose t is a number such that t < −3. Compute|4t + 5|.The important point when dealing the absolute values: To simplifythem, you must have enough information to deduce the sign to thequantity under consideration.

Sometimes, in mathematics, we are taking absolute values of sym-bolic quantities and nothing is known about the sign of the quantity.This occurs when we want to graph an expression or solve an equa-tion involving absolute values. In this case, removal of the absolutevalue takes on a conditional flavor; to do this, we simply invoke thedefinition of absolute value.


Example 1.4. Remove the absolute value in |4x − 2|.This ‘case analysis’ is an important skill that you should have. Practiceon the following two exercise. Solve and simplify completely beforelooking at the solutions.

Exercise 1.8. Remove the absolute value in |3x − 12|.Exercise 1.9. Remove the absolute values in the expression |2+3x|.Hopefully, you get the idea.

This kind of analysis can be carried out to various degrees of successinvolving arbitrarily complex expressions.

Regardless of the complexity of the expression, we can always removethe absolute value. For example, we can write |x sin(x2)| as

|x sin(x2)| ={

x sin(x2) if x sin(x2) ≥ 0−x sin(x2) if x sin(x2) < 0


However, to continue the simplification, we basically need to solve thecomplex inequality x sin(x2) ≥ 0. (This is not as hard as it mightseem.)

The absolute value interacts with multiplication and division nicely.Here are some basic properties of absolute value.

Properties of Absolute ValueLet a and b be numbers, then

1. | − a | = | a | and 2. | ab | = | a | | b | and∣∣∣ a

b

∣∣∣ =| a || b |

Equation (1) simply says that −a and a are the same distance awayfrom zero.

Illustration 1. Here are some examples of these properties.(a) |5x| = |5| |x| = 5|x|.(b) | − 7w| = | − 7| |w| = 7|w|.


(c) | − z| = |z|(d) |w4| = |w| |w| |w| |w| = |w|4.(e) |an| = |a|n, as demonstrated in the previous example.

(f)∣∣∣∣−x

9

∣∣∣∣ =|x|9

.

Exercise 1.10. Simplify each of the following.

(a) | − 56a | (b) | (−1)nx | (c)∣∣∣∣ x

−10

∣∣∣∣When we deal with solving equations and inequalities, which is takenup in Lesson 7 and Lesson 8, we’ll enumerate other useful propertiesof the absolute value.

• Distance between two numbersGiven two numbers a and b, what is the distance between these num-bers? Let’s have a notation of

d(a, b) = distance between a and b.


For example, take a = 5 and b = 9. It is quite within your experienceto agree with me that d(5, 9) = 4 ‘units’; that is, the number 5 andthe number 9 are 4 units apart. We reasoned as follows: 9 is greaterthan 5 and so

d(5, 9) = larger minus smaller = 9 − 5 = 4.

What now is d(−5,−12)? Again, we first discern the larger of the twonumbers. In this case, −12 < −5. Thus,

d(−5,−12) = larger minus smaller = −5 − (−12) = −5 + 12 = 7.

Thus, d(−5,−12) = 7.

As a general rule,

d(a, b) = larger minus smaller of a and b.

But this is a rather silly formula; let’s get fancier to confuse you!


Notice that if a and b are numbers, then from the definition of absolutevalue,

|a − b| ={

a − b if a − b ≥ 0−(a − b) if a − b < 0

This can be rewritten as

|a − b| ={

a − b if a ≥ b

b − a if a < b

If you stare at this last equation long enough, you can see that

|a − b| = larger minus smaller of a and b.

The distance between two numbers: Let a and b be numbers,then the distance between a and b is given by

d(a, b) = |a − b|. (8)

This is a powerful and important interpretation of the absolute value.


Exercise 1.11. How far is 34 away from −5.3?

Example 1.5. What can you say about the number x if a clue to itsidentity is |x − 3| < 1?

Exercise 1.12. What can be said about the number x if it is knownthat |x − 10| < 3?

Exercise 1.13. What can be said about the number x if it is knownthat |x + 4| < 3? (Hint : |x + 4| = |x − (−4)|.)Here is the other type of inequality, see if you can interpret it.

Exercise 1.14. What can be said about the number x if it is knownthat |x − 1| > 6?

• The Midpoint between two NumbersFinding the number exactly halfway between two others is a funda-mental skill. Let x1 < x2 be numbers, and let x be the midpoint


between them. Thus,

x1 < x < x2and,

d(x1, x) = 12d(x1, x2) (10)

The last equation reflects the facts that x is halfway between x1 andx2.

Now we use the distance formula, equation (8), to represent each ofthe above distances in (10).

d(x1, x) = |x1 − x|= x − x1 � since x1 < x and (8)

d(x1, x2) = |x1 − x2|= x2 − x1 � since x1 < x2 and (8)


Substitute these into equation (10) to obtain

Given: d(x1, x) = 12d(x1, x2)

Substitute: x − x1 = 12 (x2 − x1)

Solve for x: x = x1 + 12x2 − 1

2x1

x = 12x1 + 1

2x2

x =x1 + x2

2We have calculate the midpoint between two numbers:

Midpoint formula for the Real Line:Let x1 and x2 be two numbers, then the number x that ishalfway between x1 and x2 is

x =x1 + x2

2(11)


Or, in other words, the midpoint between two numbers x1 and x2 istheir (arithmetic) average.

Let’s illustrate the calculations using a quiz format.

Quiz. Answer each of the following. Passing is 100%.1. What number is midway between 2 and 9?

(a) 3.5 (b) 4.5 (c) 5.5 (d) n.o.t.2. What number is midway between −12 and 34?

(a) 11 (b) 13 (c) 23 (d) n.o.t.3. Is the point halfway between two rational numbers a rational

number?(a) Yes (b) No

4. Is the point midway between two irrational numbers an irra-tional number?(a) Yes (b) No

5. Is the point midway between two integers an integer?(a) Yes (b) No

EndQuiz


1.4. Interval Notation

In mathematics we deal with intervals of numbers. The most imme-diate application of intervals is that they are used to describe thesolution sets to inequalities, but this will not happen in a disciplinedway until Lesson 8.

In this section we introduce important notation for intervals.

Let’s make a series of definitions. Let a and b be numbers with a < b.

( a, b ) = { x | a < x < b } � an open interval

[ a, b ) = { x | a ≤ x < b } � closed on left, open on right

( a, b ] = { x | a < x ≤ b } � open on left, closed on right

[ a, b ] = { x | a ≤ x ≤ b } � a closed interval


These are intervals of finite length. We also have intervals of infinitelength.

( a,∞ ) = { x | x > a }[ a,∞ ) = { x | x ≥ a }(−∞, a ) = { x | x < a }(−∞, a ] = { x | x ≤ a }

This point marks the end of this lesson. The previous section is rathersketchy at this time and will be revised and extended some time inthe near future.

If you want to continue with Lesson 2 you are most welcome to. Theanswer section follows so bail out now.

Solutions to Exercises

1.1. Take the last digit decrease by one and post fix infinity many9’s.

423.43537 = 423.435369999999999999 . . .

Exercise 1.1.

Solutions to Exercises (continued)

1.2. Answer:

56.0 = 55.9999999999999999999 . . .

Exercise 1.2.


1.3. Let N = 3.45454545454545 . . . . Multiply this by 100 to obtain100N = 345.45454545 . . . . Now subtract the two

99N = 100N − N = 345.45454545454545 . . . − 3.45454545454545 . . .

= 345 − 3= 342

Thus,

N =34299

.

which can be reduced to

N =3811

.

Or, N = 3 511 if you prefer it written this way. Exercise 1.3.


1.4. Let N = 4.177777777777777 . . . .

N = 4.1777777777777777777 . . .

10N = 41.7777777777777777777 . . .

100N = 417.7777777777777777777 . . .

Subtract the second equation from the third to obtain

90N = 417 − 41 = 376

Thus,

N =37690

But wait, time out, there is a common factor present in the abovesolution. A better answer is then,

N =18845

Exercise 1.4.


1.5. The inequality is nonsense! From the interpretations just dis-cussed, 1 ≤ x ≤ −2 means 1 ≤ x and x ≤ −2. In words, x is greaterthan (or equal to) 1 and x is less than (or equal to) −2. Now we mustask ourselves: What number is bigger than 1 and less than −2? TheAnswer : There is no such number!

Inequalities such as 1 ≤ x ≤ −2 are often written by students whohave an imperfect understanding of the concepts and notations theyare using. I hope you are not one of these students. Exercise 1.5.


1.6. Since w < 4, we deduce that w − 4 < 0. From (7) we then have

|w − 4| = −(w − 4) � since w − 4 < 0.

= 4 − w.

Exercise 1.6.


1.7. You should have used the same reasoning as in Example 1.3.You must investigate the sign of 4t + 5:

t < −3 =⇒ 4t < −12=⇒ 4t + 5 < −12 + 5=⇒ 4t + 5 < −7

Thus, 4t + 5 < −7 < 0; that is, 4t + 5 is a negative number. So

|4t + 5| = −(4t + 5) � from (7)

• Question. Suppose I had asked to you calculate |4t+15|. What canbe said about the sign of 4t + 15 and the value of |4t + 15|?

Exercise 1.7.


1.8. Simply apply the definition:

|3x − 12| ={

3x − 12 if 3x − 12 ≥ 0−(3x − 12) if 3x − 12 < 0

This can and should be refined: 3x − 12 ≥ 0 is equivalent to 3x ≥ 12,which, in turn, is equivalent to x ≥ 4. Thus,

|3x − 12| ={

3x − 12 if x ≥ 4−(3x − 12) if x < 4

One last refinement,

|3x − 12| ={

3x − 12 if x ≥ 412 − 3x if x < 4

That seems routine! Exercise 1.8.



|2 + 3x| ={

2 + 3x if 2 + 3x ≥ 0−(2 + 3x) if 2 + 3x < 0

This can and should be refined: 2 + 3x ≥ 0 is equivalent to 3x ≥ −2,which, in turn, is equivalent to x ≥ −2/3. Thus,

|2 + 3x| =

2 + 3x if x ≥ −23

−(2 + 3x) if x < −23


|2 + 3x| =

2 + 3x if x ≥ −23

−2 − 3x if x < −23

Exercise 1.9.


1.10. Solutions:(a) | − 56a| = 56|a|(b) |(−1)nx| = |(−1)n| |x| = | − 1|n |x| = |x|.(c)

∣∣∣∣ x

−10

∣∣∣∣ =|x|10

.

Exercise 1.10.


1.11. The symbolic answer is d(34,−5.3) = |34− (−5.3)|. This eval-uates to

d(34,−5.3) = |34 − (−5.3)| = |34 + 5.3| = 39.3

Thus,d(34,−5.3) = 39.3

Exercise 1.11.


1.12. The inequality |x − 10| < 3 states that x is less than 3 unitsaway from the number 10.

Quiz. Where does that put x? Between(a) 8 and 9 (b) 7 and 12 (c) 7 and 13 (d) 8 and 13

This reasoning is the beginning of solving absolute inequalities. Morein Lesson 8. Exercise 1.12.


1.13. The expression |x− (−4)|, from equation (8), is interpreted asthe distance between x and −4; or it is the distance x is away from−4.

The inequality |x − (−4)| < 3 then states that x is less than 3 unitsaway from the number −4.

Quiz. Where does that put x? Between(a) −7 and −4 (b) −6 and −1 (c) −4 and −1 (d) −7 and −1

Exercise 1.13.


1.14. The inequality |x − 1| > 6 can be interpreted as ‘x is morethan 6 units away from the number 1.’ Exercise 1.14.

Solutions to Examples

1.1. We use a little algebraic notation. Let

N = 453.123123123123123 . . .

Based on the pattern of the repeating decimal, multiply both sides ofthe equation by 1000

N = 453.123123123123123 . . . (S-1)so

1000N = 453123.123123123123123 . . . (S-2)

Now subtract equation (S-1) from (S-2) to obtain,

999N = 452123 − 453 = 452670thus,

N =452670999

=150890333

(You did see that common factor of 3 didn’t you?)

Solutions to Examples (continued)

Answer :

453.123123123123123 · · · = 150890333

Example 1.1.


1.2. We reason as follows.

Compute |a − 5|. To remove the absolute values, we must know thesign of a − 5. It is given that a > 5; this implies a − 5 > 0. Now sincea − 5 is positive, we deduce from the definition of absolute value,equation (7), that

|a − 5| = a − 5 � since a − 5 > 0

Now compute |5 − a|. To remove the absolute values, we must knowthe sign of 5 − a. It is given that a > 5, and so 5 − a < 0. Thus, from(7), we see that

|5 − a| = −(5 − a) = a − 5 � since 5 − a < 0.

Example 1.2.


1.3. Here we are given s > 6. To compute the |2s − 3| we need toknow the sign of 2s − 3. We reason as follows.

s > 6 =⇒ 2s > 12=⇒ 2s − 3 > 12 − 3=⇒ 2s − 3 > 9

Thus, 2s − 3 > 9, in particular, 2s − 3 > 0. Thus,

|2s − 3| = 2s − 3 � from (7)

Example 1.3.



|4x − 2| ={

4x − 2 if 4x − 2 ≥ 0−(4x − 2) if 4x − 2 < 0

This can and should be refined: 4x − 2 ≥ 0 is equivalent to 4x ≥ 2,which, in turn, is equivalent to x ≥ 1/2. Thus,

|4x − 2| =

4x − 2 if x ≥ 12

−(4x − 2) if x <12


|4x − 2| =

4x − 2 if x ≥ 12

2 − 4x if x <12

and we’re out of here! Example 1.4.


1.5. The inequality |x − 3| < 1 can be interpreted quite simply. The|x − 3| is the distance x is away from the number 3. The statement|x − 3| < 1 says that the distance x is away from the number 3 isless than 1 unit. Or, rephrased, x is less than 1 unit away from thenumber 3.

Doesn’t this mean that x is somewhere between 2 and 4?Example 1.5.

Important Points

Important Points (continued)

That’s Right! The symbol N ⊆ Z means N is a subset of Z; this meansthat every natural number is also an integer.

The symbol ‘⊆’ means ‘subset’: We say A ⊆ B if it is true that everyelement of A is also a member of B, or p ∈ A implies p ∈ B.

Important Point


What does p ∈ Z mean? The symbol p ∈ Z means

• “p is a member of the set Z” or,

• “p belongs to the set of integers” or, more simply,

• “p is an integer.”

The symbol ‘∈’, which is the Greek letter epsilon, may be read as ‘isa member of’ or ‘belongs to’. Important Point


The product of two odd integers is an odd integer. You can see thisby example (for example, 3×5 = 15). More interesting, can you proveit? Try to do so using the representations of an odd and even integergiven in (3); a formal proof requires some algebra. :-)

Important Point


What is wrong with the rational number − 812? Answer : It’s not re-

duced to its lowest terms! You did see that didn’t you?

Of course, − 812 = − 2

3 . We don’t want to present fractions that arenot in reduced form! Important Point


The rational number system is a closed number system with respect toaddition, subtraction, multiplication, and division. This means that

p, q ∈ Q, implies p + q, p − q, pq,p

q∈ Q,

in the last case, we require q �= 0.

Is N closed with respect to addition, subtraction, multiplication, anddivision? The answer is No.

Question. With respect to which arithmetic operations is N closed?(a) Addition, subtraction, multiplication(b) Addition, subtraction(c) Addition, multiplication(d) n.o.t.

If you did not get it right the first time, and after you have found thecorrect response, think things through again and try to understandthe correctness of the answer.


Question. With respect to which arithmetic operations is Z closed?(Choose the “best alternative.”)(a) Addition, subtraction, multiplication(b) Addition, subtraction(c) Addition, multiplication(d) n.o.t.

It is good to think about such things. Important Point


Well, in that case, read on!

Important Point


The answer here is No in general. For example, the midpoint between0 and 1 (integers) is 1

2 (not an integer).

However, the midpoint between 0 and 2 (integers) is 1 (an integer).

Sometimes its true and sometimes its false. The reason the answer isNO is because the question was “Is it (always) true that the pointmidway between two integers is an integer?” Important Point


Repeat the solution to the exercise with modifications:

t < −3 =⇒ 4t < −12=⇒ 4t + 15 < −12 + 15=⇒ 4t + 5 < 3

We can say that 4t+5 < 3 which means it may be positive or negative;therefore, we cannot simplify the |4t + 15| in this case without moreinformation as to the value of t. Important Point

��mptiimenu


Lesson 2: Exponents & Radicals

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 2: Exponents & Radicals

Table of Contents2. Exponents and Radicals2.1. Integer Exponents2.2. The Law of Exponents

• How to Multiply or Divide Two Powers • How to Calcu-luate a Power of a Product or Quotient • How to Computea Power of a Power

2.3. Radicals• Properties of Radicals

2.4. Fractional Exponents

2. Exponents & RadicalsThis lesson is devoted to a review of exponents, radicals, and theinfamous “Laws of Exponents.” The student must have the skills tomanipulate exponents without error.

2.1. Integer Exponents

Let a be a number, and n ∈ N be a natural number. The symbol an

is defined asan = a · a · a · a · · · a︸︷︷︸

n factors

(1)

That is, an is the product of a with itself n times.

Sometimes, negative exponents enter into the mix. These are definedby

a−n =1an

where n ∈ N and a �= 0. (2)

Needless to say, we define a0 = 1, for all a �= 0.

Section 2: Exponents & Radicals

Thus, the symbol ak is defined for all integers k ∈ Z: for positiveintegers as in equation (1), 23 = (2)(2)(2) = 8; for negative integersas in equation (2), 2−3 = 1/23 = 1/8; and for zero, 20 = 1.

Terminology. The symbol ak is called a power of a. We say that ak

has a base of a and that k is the exponent of the power of a.

Numerical calculations offer no challenge to the student (that’s you).The more interesting case is when there are symbolic quantities in-volved; however, there is one situation involving numerics (and sym-bolics) in which some students—I’m not saying you necessarily—havea weakness. Consider the following . . .

Quiz. Suppose you wanted to square the number −3, what would bethe correct notational way of writing that?(a) −32 (b) (−3)2 (c) (a) and (b) are equivalent

To effectively manipulate expressions involving symbolics, we mustbe the masters of the Laws of Exponents—to be taken up shortly;just now, however, I want to illustrate how the definitions of ak areapplied.


Illustration 1. Here are several important illustrations of the tech-niques revolving about the definitions given in equations (1) and (2).

(a) x−6 =1x6 (b)

1y12 = y−12 (c) (ab)−1 =

1ab

(d)x4

y−4 = x4y4 (e)x4y−2

z−5 =x4z5

y2 (f)w9

s−5t−3 = w9s5t3

Illustration Notes: Look at (a) first from left-to-right. Basically, thissays that if we have a negative exponent in the numerator, we canshove the expression into the denominator by changing the sign ofthe exponent. Reading (a) from right-to-left, we see that if we havean expression in the denominator, we can lift it to the numerator bychanging the sign of the exponent. Similar comments can be made inequation (b).

Illustration (c) suggests that if an expression is grouped, thenwhen we move it to the denominator, we move the whole group.

Examples (d), (e) and (f) demonstrate how equations (1) and(2) are used in practice. Note that these expressions only involve mul-tiplication and/or division—most important!


Key Point. When moving expressions from the numerator to thedenominator (or from denominator to numerator), simply change thesign of the exponent.

Important. The last point is only valid for expressions involvingmultiplication and/or division. For example, the following shall bereferred to as an algebraic blunder!

NOT TRUE! =⇒ 1x−2 + y−2 = x2 + y2 ⇐= NOT TRUE!

Exercise 2.1. Remove all negative exponents for each of the expres-sions below.(a) e−4 (b)

1(x + y)−6 (c) (st)−(k+2)

(d)w6

x4y−7 (e)a7b−9c−1

d−4e6 (f)(x + y)6

x−1y−2


2.2. The Law of Exponents

When the exponent in (1) is a natural number, the following equationsshould be manifestly obvious:

The Law of Exponents—Junior Grade:Let a and b be numbers, and n and m be integers, then

1. anam = an+m

2. (ab)n = anbn and(a

b

)n

=an

bn

3. (an)m = anm

Let’s take each of these in turn, discuss them and illustrate them.


• How to Multiply or Divide Two PowersWhen multiplying two powers together with different bases, there isno simplification possible. For example, a4b5 cannot be simplified.

When multiplying two powers together having the same base, thegoverning law is

anam = an+m (3)

The validity of this identity simply follows from the definition. Thesecalculations assume n and m are natural numbers.

anam = a · a · a · a · · · a︸︷︷︸n factors

a · a · a · a · · · a︸︷︷︸m factors

= a · a · a · a · · · a · a · a · a · a · · · a︸︷︷︸n+m factors

= an+m.

Stare at this “proof.” If you can keep a visualization of this demon-stration in your head, then you cannot misuse this particular law.


Key Point. When multiplying powers with the same base, just addthe exponents. The rule anam = an+m is valid for both positive andnegative integers.

Illustration 2. Here is a list of quick illustrations of this first law.(a) 4943 = 412 (b) x3x5 = x8

(c) 5y−3y7 = 5y−3+7 = 5y4 (d)1

x3x−8 =1

x−5 = x5

(e) (s + 1)4(s + 1)−3 = (s + 1) (f) x4y3x−1y−4 = x3y−1 =x3

y

The illustrations above all had numerical values for the exponents.One can apply this first law even when the exponent is symbolic.

Illustration 3. Study these examples.(a) 4n4n+1 = 42n+1

(b) (x + 1)3n−1(x + 1)4n+3 = (x + 1)7n+2

(c) w5ks4mw3k+1s5 = w8k+1s4m+5


In each case, the base of the exponentials is the same, so we just addthe exponents.

Exercise 2.2. Combine exponents as appropriate using law #1. Firstsolve, write down your answers using good notation, then look at theanswers. (A score of 100% is expected.)(a) x4x4y2 (b) (s + 1)12(s + 1)(s + 1)3 (c) 23 y3w4w8y2

(d) 4249 (e) x4m+3x5m−1 (f) y3jy12j+m

(g) a6a−3 (h) (ab)7(ab)4(ab)−2 (i)x5y9y−2

x2y3

Now consider an/am in the case of m > n. This means that there aremore factors of a in the denominator than the numerator. The samecancellation goes on, but there factors of a left over in the denominatorwhen all the cancellation is over. Thus,

an

am=

1am−n

m > n.


Illustration 4. Simplify each.

(a)x5

x12 =1

x12−5 =1x7

(b)43t(s + 2)4

t4(s + 2)15=

43t3(s + 2)11

(c)(st)7

12s12t4=

s7t7

12s12t4=

112

s7

s12

t7

t4=

112

1s5

t3

1=

t3

12s5

The last example can be simplified in fewer steps; these additionalsteps were include for clarity.

The formula anam = an+m can be read from right-to-left as well.Below are some examples that would suggest when it would be ad-vantageous to use the first law in this way.

Illustration 5.(a) Simplifying exponentials having a negative sign in the base.

1. (−x)3 =((−1)x

)3 = (−1)3x3 = −x3

2. (−x)4 =((−1)x

)4 = (−1)4x4 = x4

3. (−2w)3 = (−2)3w3 = −8w3


4. (−3z)−2 = (−3)−2z−2 =1

(−3)21z2 =

19

1z2 =

19z2

(b) Here are some abstract versions of the same type. Let n ∈ Z bean integer.

1. (−x)2n =((−1)x

)2n = (−1)2nx2n = x2n. For any integern, 2n is an even integer. The number −1 raised to an evenpower is 1!

2. (−t)2n+1 =((−1)t

)2n+1 = (−1)2n+1t2n+1 = −t2n+1. Forany integer n, 2n + 1 is an odd integer. The number −1raised to an odd power is −1.

This kind of simplification is seen all the time in mathematics. Learnhow to master it.

Exercise 2.3. Simplify each of the following.(a)

(−(w + 1))13 (b) (−3w)4 (c) (−2s)−3

(d) (−u)2n (e) (−p)2n−1


Study these illustrations and exercises. See if you can get a “feel” forthe pattern.

Always strive to develop good algebraic techniques. Examples of goodtechniques are these tutorials. Don’t be lazy. Think of me sitting heretyping out all these examples and exercises. If I can sit here typing outexamples of good algebraic notation and techniques, can’t you take alittle time out and do the same?

The Cancellation Law. The process of combining ratios of exponen-tials having the same base can be greatly accelerated by the followingobservations:

1. If n > m (that is, the exponent in the numerator is greater thanthe exponent in the denominator), then

an

am= an−m;


2. If n < m, (that is, the exponent is the denominator is greaterthan the exponent in the numerator), then

an

am=

1am−n

.

When Law #2 is viewed this way, it is referred to as a CancellationLaw.

Here is a ‘proof by example.’

x5

x2 = x5x−2 = x5−2 = x3

Or, if you have a quick mind, you can write

x5

x2 = x5−2 = x3, or justx5

x2 = x3

Here, the numerator has the greatest exponent, so we subtract theexponent in the denominator from the exponent in the numerator,and leave the result in the numerator.


Now consider this calculation:x3

x12 = x3x−12 = x3−12 = x−9 =1x9 .

This process can be accelerated to

x3

x12 =1

x12−3 =1x9 , or to just

x3

x12 =1x9 ,

The exponent in the denominator is greater than the one in the numer-ator, we subtract the exponent in the numerator from the exponentin the denominator, and leave the result in the denominator.

Can you latch onto the visual feel of this operation?

Illustration 6. Consider the following. Simplify each.

(a)x7

x4 = x7−4 = x3.

(b)12(w + 3)23

(w + 3)3= 12(w + 3)20.


(c)x5y7

y3 = x5y4.

(d)x5

x12 =1

x12−5 =1x7

(e)43t(s + 2)4

t4(s + 2)15=

43t3(s + 2)11

(f)s7t7

12s12t4=

112

s7

s12

t7

t4=

112

1s5

t3

1=

t3

12s5

Illustration Notes: The last example is meant to delineate all the for-mal steps that go into the reasoning process. Of course, it can besimplified in fewer steps; these additional steps were included for clar-ity.

Look over each of these illustrations. They all involve multipli-cation and division. This is the only time you can apply the laws ofthe exponents!

Some expressions above, you might say, do involve additions!What do you have to say about that! I say, that these additions areenclosed with parentheses. In this case, the calculation is treated as


a single algebraic object or unit. That object is related to the rest ofthe expression by multiplications and/or divisions.

As you can see, this is a process of cancellation.Exercise 2.4. Try a few simplifications on your own. Use good no-tation, be neat and be correct, and . . . be quick and efficient.

(a)4x5y6

x3y9z2 (b)w(u + v)9

w5(u + v)3(c)

x5y5z5

5x4y5z12

• How to Calculate a Power of a Product or QuotientThe second Law of Exponents states that

(ab)n = anbn.

This tells us how to compute a power, n, of the product of two num-bers, a and b. The validity of the first is apparent when you simplywrite out the meaning of these symbols.


We give a demonstration for the case in which the exponents arenatural numbers:

(ab)n = (ab) · (ab) · (ab) · (ab) · · · (ab)︸︷︷︸n factors

� defnition

= a · a · a · a · · · a︸︷︷︸n factors

b · b · b · b · · · b︸︷︷︸n factors

� rearrange factors

= anbn. � definition

This formula is valid for negative exponents as well. Let’s look anexample with a negative exponent, and ‘prove’ the validity of (ab)n =anbn by example:

(ab)−3 =1

(ab)3� defn of neg. exp.

=1

a3b3 � Law #2

= a−3b−3� defn of neg. exp.


A Power of a Product : A power of the product of is theproduct of the powers of each. In symbols,

(ab)n = anbn.

Using the formula (ab)n = anbn from left-to-right is called expandingthe expression. Here are some examples.

Illustration 7. Expand the following.(a) (xy)4 = x4y4

(b) (4x)2 = 42x2 = 16x2

(c) 4(x(y + 1)

)23 = 4x23(y + 1)23

(d)(a(3b + 5)(4c − 3)

)17 = a17(3b + 5)17(4c − 3)17

(e) (4x)−2 = 4−2x−2 =142

1x2 =

116x2

(f)(st)4

s3 =s4t4

s3 = st4 (by Cancellation Law).

(g) a6b7(ab)−3 = a6b7a−3b−3 = a6−3b7−3 = a3b4


The formula (ab)n = anbn can be read from right-to-left as well. Inthis case, we are combining.

Illustration 8. Combine each of the following..(a) 12x12(y + 1)12 = 12 (x(y + 1))12

(b) a4b4c4 = (abc)4

(c) dmpmsm = (dps)m

(d)1

x9y9 =1

(xy)9

(e)−3x8y8

(xy)10=

−3(xy)8

(xy)10=

−3(xy)2


Now look of the ‘quotient’ version of the same law.(a

b

)n

=(a

b

)·(a

b

)·(a

b

)·(a

b

)· · ·

(a

b

)︸︷︷︸

n factors

� defnition

=a · a · a · a · · · ab · b · b · b · · · b︸︷︷︸

n factors

� rearrange factors

=an

bn. � definition

Keep a mental visualization of this ‘proof.’ Thus,

A Power of a Quotient : A power of the quotient of is thequotient of the powers of each. In symbols,(a

b

)n

=an

bn.


Illustration 9. Now let’s look at a few examples.

(a)(

x

y

)4

=x4

y4

(b)(

s

s3 + 1

)13

=s13

(s3 + 1)13

(c)(

xy

x + y

)3

=(xy)3

(x + y)3=

x3y3

(x + y)3

(d)[

x

x2 + 1

]−10

=x−10

(x2 + 1)−10 =(x2 + 1)10

x10 .

Or, we can combine, instead of expanding.

(a)x4

y4 =(

x

y

)4

(b)s13

(s3 + 1)13=

(s

s3 + 1

)13

Here’s a few for you. Be neat. Use good notation. Be organized.


Exercise 2.5. Expand each of the following:(a) (ab)5 (b) (5x)3 (c)

(x(x + 1)

)9

(d)(w

t

)6(e)

[s(s + 2)

w

]7

(f)[

u + v

u(4u + v)

]−4

Exercise 2.6. Expand and combine, eliminating all negative expo-nents.

(a) x4(xy)−2 (b)(st)−4

s3 (c) x6(

x

y

)−3

• How to Compute a Power of a PowerStudents mix this property of exponents with the first Law of Expo-nents. Here we want to calculate a power of a power. The third Lawof Exponents states that

(an)m = anm.


If you understand the origin behind these two laws, then you cannotget mixed up yourself. (We save that for someone else!) Take a ganderat the ‘proof.’ From the definition, we have

(an)m = an · an · an · an · · · an︸︷︷︸m factors

=

n factors︷︸︸︷(a · a · · · a) ·

n factors︷︸︸︷(a · a · · · a) · · ·

n factors︷︸︸︷(a · a · · · a)︸︷︷︸

m groups

= a · a · a · a · a · · · a︸︷︷︸nm factors

= anm

In the last step, we had m groups of n factors of a. The total numberof factors of a we have then is nm. This law is valid for negativeexponents as well.


Illustration 10. Here are a sample of problems.(a) (Skill Level 0) (x5)3 = x15, (x−5)3 = x−15, (x5)−3 = x−15,

(x−5)−3 = x15. (All possible combinations of signs.) :-)(b) Simplify (x4y6)3.

Solution:(x4y6)3 = (x4)3(y6)3 � Law #2

= x12y18� Law #3

Observe the use of the parentheses here. We want to protectourselves against error. For example, if we raise x4 to the 3rd

power, we denote that by (x4)3, which, in turn, simplifies to x12.

(c) Simplify(

x5

y3

)4

.

Solution: (x5

y3

)4

=(x5)4

(y3)4� Law #2

=x20

y12 � Law #3


Notice the use of the parentheses in, for example (x5)4. Withoutthose parentheses, a mindless application of Law #2 would yieldx54—this may be misinterpreted as the 54th power, not as the20th power as it should.

(d) Simplify—eliminate negative exponents:(x−4(y − 3)2

)−3.Solution:(

x−4(y − 3)2)−3 = x12(y − 3)−6

� Law #3

=x12

(y − 3)6� defn of a−n

(e) Simplify[x4(s + 1)2

5w3

]2

.

Solution:[x4(s + 1)2

5w3

]2

=(x4)2

((s + 1)2

)2

52(w3)2� Law #2

=x8(s + 1)4

25w6 � Law #3


Again not the judicious use of parentheses to protect oneselfagainst possible error.

Quiz. Which of the following is a correct simplification to(x4y−4z−3

)2

(a) x6y−2z−1 (b) x8y−2z−1 (c) x8y−8z−6 (d) x8y−8z−1

Before having you do a few problems, let’s elevate the third Law ofExponents to the status of a shadow box!

A Power of a Power : A power of a power of is the base raisedto the product of the two exponents In symbols,

(an)m = anm

Here is an exercise that uses all the Laws of Exponents. Use goodtechniques, be neat, and know how to justify each step.

Exercise 2.7. Simplify each of the following, removing any negativeexponents as needed.


(a)(a(a + 1)2(a + 2)3

)4 (b)(xy)6

(x3y4)2(c) (x−1y−1)−3

(d) (x4y−2)−3 (e)2s9t3

(st)5(f)

d−3p3s4

(dps)−2

2.3. Radicals

Life would be very tiresome if all we did in mathematics is to raiseone number to an integer power. Let’s strive forward and take up theconcept of a root of a number.

Let n ∈ N be a natural number and a ∈ R be a real number. The nth

root of the number a is defined as follows.Case 1: n is an odd number. In this case the nth root of a isdefined to be that number b ∈ R such that bn = a. In this case,write b = n

√a. Thus,

b = n√

a means bn = a. (4)


Case 2: n is an even number. In this case, the nth root of a isdefined to be that number b ≥ 0 such that bn = a. In this case,write b = n

√a. Thus,

b = n√

a means bn = a. (5)

Note: If a < 0, then there is no number b ≥ 0 that satisfies (5).Even roots exist only for nonnegative numbers.

The expression n√

a is called a radical and a is referred to as theradicand.

In the age of the electronic calculator, computing roots is trivialpursuit—actually you aren’t computing roots you are approximatingthe roots. Therefore, we shall concentrate on the more algebraic orsymbolic properties of radicals. But first, let’s take a look at somesimple examples and accompanying comments.

Illustration 11. There are a few numerical examples.(a) (Case 1.) The odd root of any number exists: For example, 3

√8 =

2, since 23 = 8; and 3√−8 = −2, since (−2)3 = −8.


(b) (Case 2.) The even root of a negative number does not exist (asa real number). Thus,

√−2, and 4√−1.12345 do not exist as real

numbers. (Note: they do exist as complex numbers).(c) (Case 2.) The even root of a nonnegative number exists: For

example,√

4 = 2, since 22 = 4; and 4√

81 = 3, since 34 = 81.

Note that the even root is defined to be a nonnegative number. Here’san important point. By definition, the

√4 is that number b that sat-

isfies three conditions:

(1) b ∈ R (2) b ≥ 0 (3) b2 = 4.

The student no doubt knows that there are two numbers b that satisfy(1) and (3). The numbers 2 and −2 both have the property that theirsquare is 4; however, only 2 satisfies all three conditions. It is 2 thatwe designate as

√4; that is,

√4 = 2. The number −2 is not referred

to as the square root of 2.

Sometimes we want to access both numbers. For example, supposewe want to find all solutions to the equation b2 = 33. The correct


way of designing all solutions is b =√

33 and b = −√33, or simply as

b = ±√33.

One last comment. Odd roots always exist and have the property thatn√−a = − n

√a for n odd. (6)

This is a quite useful simplifying tool. The equalities 3√−17 = − 3

√17,

and 5√−(2x + 1) = − 5

√2x + 1 are two examples of the use of this

property.

• Properties of RadicalsWhen manipulating algebraic quantities, we often want to manipulateexpressions involving radicals.

Let’s begin with a quiz.

Quiz. Think about each of these questions before responding. Whenyou click on the correct answer, you will jump to a discussion1. Which of the following is a correct simplification of

√a + b.

(a)√

a +√

b (b) a + b (c) a2 + b2 (d) n.o.t.


2. Let a ∈ R and n ∈ N a natural number. Is it (always) true thatn√

an = a?(a) True (b) False

End Quiz.

Properties of Radicals:Let a, b ∈ R be real numbers, and n, m ∈ N be naturalnumbers, then

1. n√

an = a, if n is odd, and n√

an = |a|, if n is even.2. n

√ab = n

√a n

√b, if n is odd, or if n is even, provided

a ≥ 0 and b ≥ 0.

3. n

√a

b=

n√

an√

b, if n is odd and b �= 0, or if n even, provided

a ≥ 0 and b > 0.4. m

√n√

a = mn√

a, if m and n are both odd; otherwise, amust be nonnegative (a ≥ 0).


Let’s take a brief look at each of these in turn.

The First Property of Radicals. The first stated property isusually no problem for students who already have experience workingwith radicals. When we are taking an odd root then

n√

an = a n odd.

For example, 3√

27 = 3√

33 = 3, 3√

x3 = x, 5√

(6a − 20)5 = 6a − 20, andso on.

However, if we are taking an even root, then we must be a little morecircumspect. The basic formula is

n√

an = |a| n even.

Recall that an even root is always nonnegative; hence, the necessityfor the absolute values.

We have no problems if a is positive. The case where a < 0 is theproblem child. For example√

(−3)2 =√

9 = 3


Notice that from this computation,√

(−3)2 = 3 �= −3. Of course, 3is related to −3 by | − 3| = 3. Thus we have shown in this examplethat

√(−3)2 = | − 3| = 3.

Illustration 12. Consider the following phalanx of examples.(a)

√x2 = |x|, and 4

√(s + 1)4 = |s + 1|.

(b)√

cos2(x) = | cos(x)|.(c) If we have information related to the sign of the radicand, we

can remove the absolute values.a.

√(x2 + 1)2 = |x2 + 1| = x2 + 1, since x2 + 1 > 0.

b. Suppose s > 3, then√

(s − 3)2 = |s − 3| = s − 3, sinces − 3 > 0.

c. Suppose t < 1, then√

(t − 1)2 = |t−1| = −(t−1) = 1− t,since t − 1 < 0.

Exercise 2.8. Simplify each of the following using the Properties ofRadicals.(a) 4

√(2s + 1)4 (b)

√sin2(x2) (c) 7

√(x − 10)7

(d) 4√

(x − 3)4, given x < 3.


Not all expressions are of the form nth roots of nth powers. An im-portant variation is

n√

akn.

That is, a is raised to an integer, k, multiple of n. The evaluation ofthis root depends on whether n is odd or even and can be successfullycarried out by the application of the Laws of Exponents.

Illustration 13. Consider the following.(a) 3

√x6 = 3

√(x2)3 = x2, since x6 = (x2)3 by the Law of Exponents,

Law 3.(b)

√x8 =

√(x4)2 = |x4| = x4. We inserted absolute values because

we are extracting an even root. But since x4 ≥ 0, we can, in turn,evaluate the absolute value to x4.

(c)√

y6 =√

(y3)2 = |y3| = |y|3. Here, y3 may be positive or nega-tive so we cannot remove the absolute values without informa-tion about the sign of y.

The Second Property of Radicals. This property is veryfrequently used; care must be made not to abuse it. Property 2 states


that “a root of a product is the product of the roots, if n is odd, or ifn is even, provided all factors are nonnegative.”

The bad case is when a < 0 and b < 0 and we are taking an even rootof ab; in this case,

n√

ab �= n√

an√

b.

To see this, put n = 2, a = −1 and b = −1, to get2√

(−1)(−1) = 2√

1 = 1 �= 2√−1 2

√−1

We run into problems when we take an even root of a negative num-ber. (A no, no.) This is rather obvious, but becomes less obvious whendealing with symbolic quantities. Whenever you are tempted, for ex-ample, to simplify an expression like

√x2y, you must be very careful

not to write x√

y.

Quiz. What is the simplification of√

x2y?(a) x

√y (b) |x|√y (c) I cannot be simplified


This second property is used, in combination with the first propertyof radicals to extract perfect roots.

Illustration 14. Extract perfect roots.(a)

√x(x + 1)2 =

√x√

(x + 1)2 =√

x |x + 1| = |x + 1|√x.(b)

x4√

x4y=

x4√

x44√

y =x

|x| 4√

y. (No additional simplification is

possible unless more information is known about the sign of x.1. If it is known that x ≥ 0, then |x| = x and so

x

|x| 4√

y=

x

x 4√

y=

14√

y

2. If it is known that x < 0, then |x| = −x and so

x

|x| 4√

y=

x

(−x) 4√

y=

1− 4

√y

= − 14√

y

(c)√

x4y =√

x4√y = |x2|√y = x2√y.


(d) In this example, more detail is presented.√

a4b6 =√

a4√

b6 � Prop. Rads. #2

=√

(a2)2√

(b3)2 � Law Exp. #3

= a2 |b3| � Prop. Rads. #1

= a2 |b|3. � A Prop. Abs. Value #2

Illustration Notes: Illustration (d) is a good illustration how the var-ious properties of exponents, of radicals, and of absolute values areutilized. When you become an expert, this simplification would bejust a single step, not four steps as I have above.Property #2 can also be used to partial extraction. Let me illustratewhat I mean.

Illustration 15. Study the following examples.(a)

√x3 =

√x2x =

√x2

√x = |x|√x = x

√x. In the last equality we

were able to remove the absolute value symbols. Why?(b) 3

√w10 = 3

√w9w = 3

√w9 3

√w = w3 3

√w


The Third Property of Radicals. This property is the sameas #2, but for quotients. Just two examples are sufficient.

Example 2.1. Simplify the following:

√x8

y10 .

Example 2.2. Simplify 3√

x−18.

Exercise 2.9. Simplify each of the following utilizing the Propertiesof Radicals, the Laws of Exponents, and properties of Absolute Value.

(a)√

25x (b)√

w2s6 (c)√

x5y12

(d) 5√−w12 (e)

√a5

b6 (f)3√

x6y4√x4y6

The Fourth Property of Radicals. This law describes whata root of a root is; it’s . . . another root! Let’s work it out. We wantto understand the simplification of m

√n√

a. Let b = m√

n√

a. The char-acteristic property of b is that bm = the radicand = n

√a. Now if

bm = n√

a, then by the definition, (bm)n = the radicand = a. Thus,


bmn = a. But this is the definition of b being the mnth root of a; i.e.,b = mn

√a. We have shown that

m√

n√

a = mn√

a.

Illustration 16.(a) 3

√5√

x = 15√

x.(b) 4

√√x10 = 8

√x10 = 8

√x8x2 = 8

√x8 8

√x2 = |x| 8

√x2. (7)

• Question Argue that4√√

x9 = 8√

x9 = x 8√

x

is true. Compare this equation with (7) and explain why the absolutevalue is not needed.


2.4. Fractional Exponents

Let me finish off Lesson 2 by a (hopefully) brief discussion of frac-tional exponents. We begin by making a series of definitions.

Definition. Let a be a number.1. For n a natural number, define a1/n = n

√a, provided the root

exists as discussed in the definition of radicals.2. For n a natural number, define a−1/n =

1a1/n

.

3. If p and q are integers, define ap/q = (ap)1/q. (Of course, we areassuming q �= 0.)

Illustration 17. Consider the following conversions from radical no-tation to exponential notation and from exponential notation to rad-ical notation.(a) 3

√x = x1/3 and 6

√w2 + 1 = (w2 + 1)1/6.

(b) 5√

x−2 = x−2/5 and 3√−w−4 = − 3

√w−4 = −w−4/3 = − 1

w4/3 .


(c) s1/2 =√

s and t−1/5 =1

t1/5 =15√

t.

(d) 3√

x2 = x2/3, 2√

x5 = x5/2 and (w + 3)3/4 = 4√

(w + 3)3.

(e) x−2/7 =1

x2/7 =1

7√

x2.

Exercise 2.10. Convert each of the following to exponential notationusing the above definitions.(a) 4

√a (b) 8

√s + t (c) 4

√s3

(d) 12√

x7 (e) 3√−x2 (f) 4

√x−1

Now let’s convert from exponential to radical notation. Carefully workout the problems before you look at the solutions. Justify each stepthat you make in your solution.

Exercise 2.11. Convert to each to exponential notation.(a) 43/2 (b) 8−1/3 (c) (−4)−3/2

(d) (w + 1)2/3 (e) x−1/2 (f) s−4/5


Here’s an important point: The fractional exponent notation is en-tirely consistent with reduction of fractions to lowest terms. What doI mean by that?

Illustration 18. Reduction of Fractional Exponents.(a) x6/3 = x2 and y−12/3 = y−4.(b) w4/12 = w1/3, for w ≥ 0.(c) 8

√x2 = x2/8 = x1/4 = 4

√x.

Quiz. Think about each of the following questions before responding.1. The expression 12

√y3 is equivalent to

(a) 9√

y (b) 4√

y (c) 3√

y (d) n.o.t.2. The expression 12

√y4 is equivalent to

(a) 4√|y|3 (b) 3

√y2 (c) 3

√|y| (d) n.o.t.End Quiz.

The fractional exponent notation is entirely consistent with the Lawsof Exponents. Let’s restate this law for the record.


The Law of Exponents—Senior Grade:Let a and b be numbers, and r and s be rational numbers,and assume further that ar and as are defined. Then

1. aras = ar+s

2. (ab)r = arar and(a

b

)r

=ar

ar

3. (ar)s = ars

All the above equalities are conditional on the existence of each of theexponentials. For example, (ab)r = arbr is true provided that each ofthe exponential (ab)r, ar, and br exist. (Take a look at the second lawabove for the case a = −1, b = −1, and r = 1/2—catastrophe!)

Exercise 2.12. Give examples of situations in which Law #1 andLaw #3 are not valid.


In other words, all the techniques outlined and illustrated above ap-plies to fractional exponents.

Illustration 19. Here are some examples that illustrate the Law ofExponents—Senior Grade.

1. Law 1: Different exponent, same base; just add exponents.a. x3/2x5/2 = x3/2+5/2 = x8/2 = x4.b. w2/3w−1/6 = w2/3−1/6 = w1/2

c.x1/2y3/2

4x−1/2y2 =x1/2+1/2

4y2−3/2 =x

4y1/2 =x

4√

y.

d. x7/2 = x3+1/2 = x3x1/2 = x3√x.2. Law 2: Same exponent, different base.

a. (xy)2/3 = x2/3y2/3.b. (st)−3/4 = s−3/4t−3/4. Here, we are taken even roots of s

and t; therefore, we must assume s > 0 and t > 0.3. Law 3: Nested exponentials, just multiply exponents.

a. 45/2 = (41/2)5 = (2)5 = 32. Whereas 45/2 = (45)1/2 =(1024)1/2 = 32. Naturally, the first computation was quite


a bit less complicated; the idea is to choose the order ofcalculation that is easiest.

b. (x3)2/3 = x2.c. (w−3/4)−3 = w9/4.d. Nested radicals are the same as nested exponents, let Ex-

ponent Law #3 handle it.3√

4√

x = (x1/4)1/3 = x1/12 = 12√

x.

Exercise 2.13. Let n be an even positive integer. Simplify (an)1/n.

Here are some routine practice exercises for you solve.

Exercise 2.14. Simplify each of the following using the Law of Ex-ponents.

(a) w−3/2y2/3w3y1/2 (b)x2/3

x3 (c) (4x)−5/2(4x)2

(d) (xy)3/2x4 (e)a1/2b3/2

(ab)1/2 (f) (x3y2/3)4/3


This is the end of Lesson 2! If you have managed to work throughit, you have my congratulations! DPSIf you are not too exhausted, continue on to Lesson 3. See you there!


2.1. Passing is 100%!

(a) e−4 =1e4 .

(b)1

(x + y)−6 = (x + y)6.

(c) (st)−(k+2) =1

(st)k+2 .

(d)w6

x4y−7 =w6y7

x4 .

(e)a7b−9c−1

d−4e6 =a7d4

b9ce6 .

(f)(x + y)6

x−1y−2 = xy2(x + y)6.

Exercise 2.1.


2.2. And the answers with any further comments are(a) x4x4y2 = x8y2

(b) (s + 1)12(s + 1)(s + 1)3 = (s + 1)16

(c) 23 y3w4w8y2 = 23 y5w12

(d) 4249 = 411

(e) x4m+3x5m−1 = x9m+2

(f) y3jy12j+m = y15j+m

(g) a6a−3 = a3

(h) (ab)7(ab)4(ab)−2 = (ab)7+4−2 = (ab)9

(i)x5y9y−2

x2y3 = x5y9y−2x−2y−3 = x5−2y9−2−3 = x3y4

Add the exponents of all exponentials having the same base; bringfactors in the denominator by changing the sign on the exponent,then add exponents as appropriate. Exercise 2.2.


2.3. Solutions:(a)

(−(w + 1))13 = (−1)13(w + 1)13 = −(w + 1)13.

(b) (−3w)4 = (−3)4w4 = 81w4.

(c) (−2s)−3 = (−2)−3s−3 =1

(−2)31s3 =

1−8

1s3 = − 1

8s3 .

(d) (−u)2n = (−1)2nu2n = u2n.(e) (−p)2n−1 = (−1)2n−1p2n−1 = −p2n−1.

In the last problem, for any given integer n, 2n − 1 is an odd integer;hence, (−1)2n−1 = −1 Exercise 2.3.


2.4. Solutions:

(a)4x5y6

x3y9z2 =4x2

y3z2 .

(b)w(u + v)9

w5(u + v)3=

(u + v)6

w4 .

(c)x5y5z5

5x4y5z12 =x

5z7 .

Exercise Notes: In the case of solution (c), we automatically simplified

y5/y5 to 1; i.e.,y5

y5 = y0 = 1.

Concentrate on using good techniques. Don’t be sloppy. Don’trevert to your old bad habits. Be precise in all things you do.

Exercise 2.4.


2.5. Answer : Apply Law #2.

(a) (ab)5 = a5b5

(b) (5x)3 = 53x3 = 125x3

(c)(x(x + 1)

)9 = x9(x + 1)9

(d)(w

t

)6=

w6

t6

(e)[s(s + 2)

w

]7

=

(s(s + 2)

)7

w7 =s7(s + 2)7

w7

(f)[

u + v

u(4u + v)

]−4

=(u + v)−4

u−4(4u + v)−4 =u4(4u + v)4

(u + v)4

In the last problem, the negative exponents were eliminated by movethem to the numerator or denominator, as appropriate, and changingthe signs of their exponents—standard techniques.

Did you use good notation? Were you neat? Were you organized?Exercise 2.5.


2.6. Solutions: Apply combinations of Laws #1 and #2.

(a) x4(xy)−2 = x4x−2y−2 = x2y−2 =x2

y2

(b)(st)−4

s3 =s−4t−4

s3 =1

s4t4s3 =1

s7t4

(c) x6(

x

y

)−3

= x6 x−3

y−3 =x6x−3

y−3 = x3y3

Each of us develop our own pattern of thoughts; whatever patternof thinking we develop, ultimately, it must be based on the simpleapplication of these laws of exponents.

In these solutions, I’ve included a few more steps than is really usedin practice. Exercise 2.6.


2.7. Solutions:(a)

(a(a + 1)2(a + 2)3

)4 = a4(a + 1)8(a + 2)12.

(b)(xy)6

(x3y4)2=

x6y6

x6y8 =1y2

(c) (x−1y−1)−3 = x3y3

(d) (x4y−2)−3 = x−12y6

(e)2s9t3

(st)5=

2s9t3

s5t5=

2s4

t2

(f)d−3p3s4

(dps)−2 =d−3p3s4

d−2p−2s−2 =p3p2s4s2

d3d−2 =p5s6

d

The last example requires a careful application of the Law of Expo-nents. Hope you were are careful as I was.

Did you work the problems out first? Were you neat? NOW is thetime to develop good habits . . . not during a test. Exercise 2.7.


2.9. Solutions:

(a)√

25x =√

52x =√

52√

x = 5√

x.

(b)√

w2s6 =√

w2√

(s3)2 = |w| |s3| = |w| |s|3.(c)

√x5y12 =

√x5

√y12 =

√x4x |y6| =

√x4

√x y6 = x2√x y6. Here

we have used the fact that y6 ≥ 0, hence, |y6| = y6; similarly,√x4 = |x2| = x2, since x2 ≥ 0.

(d) 5√

−w12 = − 5√

w12 = − 5√

w10w2 = − 5√

(w2)5 5√

w2 = −w2 5√

w2.

(e)

√a5

b6 =

√a5

√b6

=

√a4

√a

|b|3 =a2√a

|b|3 .

(f)3√

x6y4√x4y6

=3√

x6 3√

y4√

x4√

y6=

x2 3√

y3y

x2√

(y3)2=

x2y 3√

y

x2|y3| =y 3√

y

|y|3 .

In problem (f), no further simplification is possible without knowledgeof the sign of y. Exercise 2.9.


2.10. Solutions:(a) 4

√a = a1/4.

(b) 8√

s + t = (s + t)1/8.(c) 4

√s3 = s3/4.

(d) 12√

x7 = x7/12.(e) 3

√−x2 = − 3√

x2 = −x2/3.(f) 4

√x−1 = x−1/4.

Conversion to exponential must be automatic. Exercise 2.10.


2.11. Solutions:(a) 43/2 =

√43 =

√64 = 8.

(b) 8−1/3 =1

81/3 =13√

8=

12.

(c) (−4)−3/2 =√

(−4)3 =√−64. Oops! An even root of a negative

number! Therefore, (−4)−3/2 does not evaluate to a real number;It evaluates to a complex number. Did you panic?

(d) (w + 1)2/3 = 3√

(w + 1)2.

(e) x−1/2 =1

x1/2 =1√x.

(f) s−4/5 =1

s4/5 =1

5√

s4.

Exercise 2.11.


2.12. Law #1. Consider the case a = −1, r = 1/2 and s = −1/2.In this case, ar+s = (−1)1/2−1/2 = (−1)0 = 1, yet ar = (−1)1/2 andas = (−1)−1/2 are not defined as a real number.

Law #3. Take a = −1, r = 1/2 and s = 2, then

(a1/2)2 �= a(1/2)(2)

The right-hand side is a1 = a = −1, whereas the left-hand side has init a1/2 = (−1)1/2, which is not a real number.

In each of the two examples, one of the expressions is not a real number. . . what’s where we ran into trouble. The Laws are valid wheneverall exponentials are defined. Exercise 2.12.


2.13. This is another exception to Law #3.

(an)1/n = n√

an = |a|by the (1) of the property of radicals. Thus, if a < 0, then

(an)1/n = |a| �= a.

Doesn’t this contradict the third law of exponents? The premise ofthe laws of the exponents is that ar and as are both defined. In thissituation, for a < 0, a1/n is not defined since we are taking an evenroot of a negative number; therefore, the Laws of the Exponents doesnot guarantee the validity of Law #3. Exercise 2.13.


2.14. Solutions:

(a) w−3/2y2/3w3y1/2 = w(−3/2+3)y(2/3+1/2) = w3/2y7/6.

(b)x2/3

x3 =1

x(3−2/3) =1

x7/3 .

(c) (4x)−5/2(4x)2 = (4x)(−5/2+2) = (4x)−1/2 =1

(4x)1/2 =1

2√

x.

(d) (xy)3/2x4 = x3/2y3/2x4 = x(3/2+4)y3/2 = x11/2y3/2.

(e)a1/2b3/2

(ab)1/2 =a1/2b3/2

a1/2b1/2 = b(3/2−1/2) = b.

(f) (x3y2/3)4/3 = (x3)4/3(y2/3)4/3 = x4y8/9.Exercise 2.14.


2.1. Solution:√x8

y10 =

√x8√y10

� Prop. Rads. #3

=x4

|y5| � Prop. Rads. #1 (S-1)

=x4

|y|5 � A Prop. Abs. Value #2

Example Notes: In equation (S-1), no absolute values were needed forthe x4 since x4 ≥ 0; however, it is possible for y5 to be negative, hencethe absolute values were needed.

Example 2.1.


2.2. Solution:

3√

x−18 = 3

√1

x18 � defn of a−n

=1

3√

x18� Prop. Rads. #3

=1

3√

(x6)3� Law Exp. #3

=1x6 � Prop. Rads. #1

Example 2.2.

Important Points


That’s Right! The correct answer is (b).

The square of −3 = (−3)2 = (−3)(−3) = 9

First choice or second? If you chose (a) as your first choice, you havea weakness in this area. Some students erroneously write −32 whenwhat they really mean is (−3)2. A mathematician would interpret theexpression −32 as −(3)(3) = −9; that is the correct interpretation ofthe notation −32 is the negative of the number 3 squared.

You must be careful about writing −32 when, in reality, you mean(−3)2. You and the one grading your paper might have a differencein opinion about the meaning of the symbol—guess who wins theargument.

Therefore, when raising a number that has a negative sign, alwaysenclose the number, including negative sign, with parentheses. Thus,cube of the number −2 is (−2)3 = (−2)(−2)(−2) = −8.


Similar comments can be made about symbolic numbers: The cube ofthe number −x is (−x)3 not −x3. More on this later.

Important Point


Good Choice! The correct response is ‘n.o.t.’, which means ‘none ofthese.’

A common error students make when (trying) to manipulate radicalsis to essentially write (on a test paper, for example)

NOT TRUE! =⇒ √a + b =

√a +

√b ⇐= NOT TRUE!

But this is an algebraic blunder! Don’t do that! The root of a sum oftwo quantities in not equal to the sum of the roots. A simple examplewill illustrate

Symbolic quantities such as a and b represent numbers. Equations youwrite involving symbolic quantities must be true when the symbolsare replaced with numbers. To see that the root of a sum is not equalto the sum of the roots, just give a and b appropriate values.

Take a = 9 and b = 16. Thus,√

9 + 16 =√

25 = 5 �= 7 = 3 + 4 =√

9 +√

16


Look at the extreme left and right sides of this nonequation. Whatdo you see? √

9 + 16 �=√

9 +√

16

That is, √a + b �= √

a +√

b

Don’t make this mistake ever again! Important Point


Correct Again! It is not always true that n√

an = a. Sometimes theradical n

√an equals a and sometimes it doesn’t equal a. Here are some

examples.

Example 1. Situation where n√

an = a is false. Think of a = −1 andn = 2, then

2√

an = 2√

(−1)2 = 2√

1 = 1 �= −1 = a.

Example 2. Situation where n√

an = a is true. Think of a = 1 andn = 2, then

2√

an = 2√

12 = 2√

1 = 1 = a.

Sometimes it true, sometimes its false. When trying to simplifying asymbolic expression like n

√an, it turns out that we need to know the

sign of the number a.

Continue now the discussion following this quiz for a definitive expla-nation of how to simplify radicals like n

√an. Important Point


Way to go, mate! Because we are taking the square root of x2y, it isan implicit assumption that x2y ≥ 0. Since x2 ≥ 0 regardless of thevalue of x, we deduce y ≥ 0.

Now by Law #2 of the Properties of Radicals we have√x2y =

√x2√y

= |x|√y � Law #1

Important Point


The original expression was√

x3, from which we deduce that x3 ≥ 0;hence, x ≥ 0. Thus, |x| = x since x ≥ 0. ’Nuff Said!

Important Point


Solution:4√√

x9 = 8√

x9

= 8√

x8x

= 8√

x8 8√

x

= |x| 8√

x

= x 8√

x

The last step needs some comment. Because we started with the ex-pression 8

√x9, we conclude that x9 ≥ 0. But x9 ≥ 0 implies that x ≥ 0,

because we are dealing with an odd power of x.

Finally, x ≥ 0 implies |x| = x. What was simple!

Awareness of the signs of the quantities is often essential to a success-ful simplification. Important Point


Way to go! Convert to exponential notation, reduce fractions, thenreturn to radical notation.

12√

y4 = y4/12 = y1/3 = 3√

y.

Important Point

��mptiimenu


Lesson 3: Basic Algebra, Part I

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 3: Basic Algebra, Part I

Table of Contents3. Basic Algebra, Part I

• You are manipulating Numbers3.1. The Basics

• The Arithmetical Operations • Terms vs Factors • Paren-theses, Brackets, and Braces • How to Negate Correctly• How to Invert Correctly

3.2. How to Add/Subtract Algebraic Expressions• Adding/Subtracing Grouped Expressions

3. Basic Algebra, Part IAlgebra is the language of mathematics, engineering and the sciences.We use algebra to express our thoughts, ideas, and to communicatewith others who understand the language of algebra.

Algebra is a language in which we can precisely pose ourselves ques-tions; Algebra is a set of tools for answering those questions.

• You are manipulating NumbersWhen we manipulate algebraic quantities, we are, in fact, manipu-lating numbers. This point must be ever kept in mind. The rules formanipulating algebraic quantities reflect the properties of the numbersystem. This is an important point. Therefore, when you manipulatesymbolic quantities in a questionable way, you must ask yourself thequestion, “Is what I have just done valid when I replace the symbolsby numbers?”

This should be your guiding principle.

Section 3: Basic Algebra, Part I

A Fundamental Guiding Principle of Algebra:“Is the algebraic manipulation that I just performed validwhen I replace the symbols by numbers?”

3.1. The Basics

In this section we take a brief survey of the arithmetical operationsand some of the properties of these operations that are exploited toperform algebraic manipulations.

• The Arithmetical OperationsLet the letters a, b, and c represent (real) numbers. As you well knowwe can add, subtract, multiply, and divide these numbers. In algebra,these operations are carried out symbolically :

1. Addition. The sum of a and b is a + b. The number a is calleda term of the expression a + b. (Of course, b is a term too.)


2. Subtraction. The difference of a and b is a − b. The number ais called a term of a − b. (Of course, b is a term too.)

3. Mulitplication. Product of a and b is ab or a · b, or sometimes,a× b. The numbers a and b are called factors of the product ab.

4. Division. The quotient of a by b isa

b, or a/b, or less frequently

(in algebra), a÷b. Division is only defined when the denominatorb = 0. The number a is the numerator, and b is the denominator.

Any combination of symbols and numbers related by the above arith-metical operations is called an algebraic expression.

• Terms vs FactorsYou should be particularly sensitive to the meaning of term and fac-tor. Roughly, a term of an algebraic expression is a symbol that isconnected to the rest of the expression by the operations of addi-tion or subtraction. A factor of an algebraic expression is a symbolthat is connected to the rest of the expression by the operations ofmultiplication.


Here are some examples of using the terminology described.

Illustration 1.(a) The number x is a factor of the expression 2x(x + 1). Both 2

and (x + 1) are factors of the expression 2x(x + 1) as well.(b) The symbol w is a term of the expression 6a+w. The algebraic

quantity 6a is also a term of the expression.(c) Consider the ratio,

xy

a + b. The symbol x is a factor of the nu-

merator. The symbol a the not a factor of the denominator, itis a term of the denominator.

(d) In the product x3(y + 1)4, x3 is a factor , as is (y + 1)4. Notethat y is not a factor of x3(y + 1)4.

It is important that you be able to recognize factors and terms ofalgebraic expression. This is often key to simplifying an expression.

Quiz. Answer each of the questions 1–8. Passing is 100%.1. How many terms does the expression 4x3 − xy + xyz2?

(a) 1 (b) 2 (c) 3 (d) 42. Is x2 a factor of x(x + 1)?


(a) Yes (b) No3. Is xy a factor of the second term of x + xy2 + xz2?

(a) Yes (b) No4. Is x2 a factor of both the numerator and denominator of the

expression(1 + y3)x3

6y3x2 ?

(a) Yes (b) No5. Is x3 a factor of both the numerator and denominator of the


6y3x2 ?

(a) Yes (b) No6. Is y3 a factor of both the numerator and denominator of the


6y3x2 ?

(a) Yes (b) No7. Is x2 a common factor to all terms of the expression x3y−4x6+

3x7/2y3?(a) Yes (b) No


8. Is xy a common factor to all terms of the expression xy3 −5x3y2 + 1.(a) Yes (b) No

EndQuiz

A Tip. Get into the habit of automatically identifying the factorsand terms of an algebraic expression. Factors that are common toboth numerator and denominator can be canceled; factors that arecommon to all terms of an expression can be factored out.

• Parentheses, Brackets, and BracesThroughout Lesson 1 and Lesson 2 I have utilized, on occasion, paren-theses to group an expression that is to be treated as a unit. It isimportant for you to acquire the ability to correctly use the groupingmechanisms of (parentheses), [brackets], and {braces}.When we want to square the quantity x+y we would use the notation(x+y)2. Without the parentheses the meaning of x+y2 has an entirelydifferent meaning.


Another example of same type is the squaring of the number −3.That would be (−3)2. The number −3 consists of two symbols: theminus sign ‘−’; and the number ‘3’. These two go together as a unit, asingle entity. That being the case, we would write (−3)2. A subset ofstudents would express the same calculation as −32, but, to someoneconversant in algebra, this has an entirely different meaning :

(−3)2 = (−3)(−3) = 9 whereas − 32 = −(3 · 3) = −9,

completely different evaluations! Some students say, “I wrote it and Iknow what I meant by −32.” That’s not the point is it? The point isto communicate with others in a way that they will understand whatyou mean without you being there to explain what you meant in anygiven situation.

Quiz. A student of mine who is in the habit of writing −32 instead of(−3)2 has the problem of writing “minus the number three squared”in algebraic notation. How will he write this down on paper?(a) −32 (b) n.o.t.


Quiz. Evaluate the expression x+x2 at x = −1. Which of the followingis the correct evaluation?(a) −2 (b) 0 (c) 2 (d) n.o.t.

Illustration 2. Here are some representative examples of grouping.When there are “nested” groups, we use brackets and/or braces.(a) x[ 1 + (x + 1)1/2 ]2.(b) (2x2)3.(c) (x + 1){ 1 + [ 1 + (x + 1)2 ]2 }2.

Exercise 3.1. To test your ability to read algebraic expressions thatcontain groupings, in Illustration 2, put(a) Put x = 3 in x[ 1 + (x + 1)1/2 ]2 and evaluate.(b) Put x = −2 in (2x2)3 and evaluate.(c) Put x = 0 in (x + 1){ 1 + [ 1 + (x + 1)2 ]2 }2 and evaluate.

Passing score is 100%.


Sometimes we write mathematics within the body of text. When wedo this, we must be careful writing the expressions. Do not write,for example, 1/2+x when you actually mean 1/(2+x). Do not writex+y/z when you mean (x+y)/z. Use parentheses, brackets, and braceswhenever there is the possibility of confusion about the meaning ofthe expression.

Exercise 3.2. Calculate the values of 1/2 + x and 1/(2 + x) for thecase of x = 1. Do they evaluate to the same numerical value? If not,what is your conclusion?

• How to Negate CorrectlyLet a be an algebraic symbol. The symbol −a is the negation of a, oradditive inverse of a. The symbol −a has the property that

a + (−a) = 0. (1)

Negation of a Positive Number : The negation of 3 is −3, the negationof 2

3 is − 23 . These are well known observations.


The Negation of a Negative Number : The negation of −6 is −(−6) = 6.The negation of −√

3 is −(−√3) =

√3. These follow from the well

known fact that a “minus times a minus is a plus.”

In general, we have the following principle:

−(−a) = a (2)

The negation of a can also be thought of as (−1) · a. That is,

−a = (−1) · a (3)

It is from this interpretation of negation that equation (2) comes.Indeed,

−(−a) = (−1)((−1)a) = (−1)(−1)a = a.

This interpretation is important when we make calculations such as(−a)5 or (−a)2k, where k ∈ Z in an integer. These kinds of evalua-tions were discussed in Lesson 2, in the section entitled The FirstExponential Law; in particular, in that section, see the illustration foradditional examples.


Exercise 3.3. Using equation (3) and the Law of Exponents, sim-plify each of the following.

(a) (−x)5 (b) (−x)2k (c)[−x3y2

z4

]3

Negating a grouped algebraic expression is always a problem for stu-dents. Here are some simple rules.

Negating a Grouped Expressions:

−(a + b) = −a − b

−(a − b) = −a + b(4)

In words, to negate the sum or difference of an expression,simply change the sign of each term.

Illustration 3. Examples of negating sum/differences of terms.(a) −(3x + 4xy) = −3x − 4xy.(b) −(−4xy + 3x2) = 4xy − 3x2.


(c) −(a − b + c) = −a + b − c.(d) Mulitple Nesting. −[−x − (4x − 2x2)] = −[−x − 4x + 2x2] =

x + 4x − 2x2 = 5x − 2x2. When you have nested expressions, itis best to work your way from the inner-most level to the outermost level of nesting.

A Common Error. Students often make sign errors because of poor useof (parentheses), [brackets], and {braces}. If a student writes −3x−4ywhen, in reality, the correct expression should have been −(3x − 4y).The first rendering, −3x − 4y does not simplify anymore whereas,−(3x − 4y) simplifies to −3x + 4y.

The next example illustrates two approaches to simplifying an expres-sion.

Example 3.1. Simplify −[x − (−4x + y)].

Exercise 3.4. Simplify each of the following expressions.(a) −(−3x + 5y) (b) −(9 − (−2x)3)(c) −(xy − (3xy − 4)) (d) [1 − (−1)3] −

[( 12

)3 − (− 12

)3]


Tip. Use (parentheses), [brackets], and {braces} liberally to “delimit”your expressions. This is especially important when dealing with theproblems of negating and subtraction. Write −(a + b) when you wantto “negate” the expression a + b; write x − (a + b) when you want tosubtract a + b from x.

• How to Invert CorrectlyWe have seen in Lesson 2 that

a−n =1an

n ∈ Z. (5)

In particular, when n = 1 we obtain,

a−1 =1a

(6)

The expression a−1 or 1/a is called the (multiplicative) inverse of a.(Of course, we are assuming a = 0.) The number a−1 has the propertythat

a−1 · a = 1 or1a

· a = 1.


In this (hopefully, short) section we discuss the basics of invertingalgebraic expressions.

Problems are encountered we try to invert fractions. Here is the guid-ing equation: (a

b

)−1=

b

a(7)

Let’s see how, operationally, this formula is used.

Illustration 4. Study the following examples.

(a)( 45x

)−1=

5x4

by (7).

(b)(x + 1

x + y

)−1=

x + y

x + 1, by (7).

Some important principles can be extracted from equation (7).


1. Inverse of an Inverse: For any a = 0,

(a−1)−1 =

(1a

)−1= a.

Thus, (a−1)−1 = a.

This equation states that a is the multiplicative inverse of a−1

and really does not represent anything new. According to theLaw of Exponents Law #3, we would simplify (a−1)−1 nor-mally by multiplying the exponents together to obtain a.

2. The inverse of a fraction. Let a = 0 and b = 0, then

1a

b

=(a

b

)−1=

b

a.

Thus,1a

b

=b

a. (8)


This equation will be used to simplify ratios of fractions—staytuned.

Here are some examples of the use of equation (8). Read the solutionscarefully and try to understand the use of equation (8).

Example 3.2. Simplify each of the following:

(a)1

x3/y2 (b)1

x/(x + y)(c)

1x/(x + y)−3 .

Illustration 5. Equation (5) can be used when negative powers otherthan −1 are involved.(a) Consider the following general result.

(a

b

)−n

=[(a

b

)−1]n

� Law Exp. #3

=[

b

a

]n

� by (7)

=bn

an� Law Exp. #2


That is, we “flip the fraction” and raise each to the power n.(a

b

)−n

=bn

an. (9)

(b) Here is a simple example:(x + y

z

)−3=

z3

(x + y)3.

(c) The numerator and denominator may themselves have powersin them. In that case we would also use Law #3 of the Law ofExponents. A particular manifestation of this formula is[ (x + y)4

z3

]−3=

z9

(x + y)12

The above calculation can be easily executed by the inclusionof an intermediate step:[ (x + y)4

z3

]−3=

[ z3

(x + y)4]3

=z9

(x + y)12


One can remove the negative exponent by inverting the fraction,then we can cube the result.

(d) Another example of the same type:[x5/4

y1/3

]−4=

[y1/3

x5/4

]4=

y4/3

x5

Now it’s your turn. Solve each of the following without error. Mysolutions may not be identical to yours. You may have to additionalalgebraic steps to verify your own answer. Try to justify each step.

Exercise 3.5. Simplify each of the following by eliminating all neg-ative exponents using equations (6), equation (8) and the techniqueof “flipping the fraction” described in equation (9).

(a)[xy + 1

x + y

]−1(b)

[ (x + y)4

x1/2y3/2

]−4(c)

[ ab−1

(a − b)−4

]−2

(d) (x3y4)−2 (e)1

x2/y4 (f)[ −x2

y3/z4

]3


3.2. How to Add/Subtract Algebraic Expressions

In this section we discuss methods of adding and subtracting algebraicexpression that do not involve operation of division. When addingand/or subtracting the terms of an algebraic expression, we combinesimilar terms.

Adding Similar Terms:To add similar terms, just add their numerical coefficients.

Illustration 6.(a) 4x3 + 6x3 = (4 + 6)x3 = 10x3. Here, we have just added the

numerical coefficients.(b) 6xy3 − 2xy3 = (6 − 2)xy3 = 4xy3.(c) 9(x2 + 1) + 6(x2 + 1) = 15(x2 + 1).

Exercise 3.6. Combine each of the following.(a) 5xyz2 + 6xyz2 − 8xyz2 (b) 6x

√x − 3

2x√

x


Parentheses are used to group similar terms together in more complexexpressions. Within each group, the problem of combining is at thesame Skill Level as the previous illustration and exercise!

Illustration 7.(a) 5x − 6y + 12x + 1

2y = (5x + 12x) + (−6y + 12y) = 17x − 11

2 y.(b) Combine 6xy + 7x + 3

√x + xy − √

x.

6xy + 7x + 3√

x + xy − √x = (6xy + xy) + 7x + (3

√x − √

x)

= 7xy + 7x + 2√

x.

Exercise 3.7. Combine each of the following expressions. As one ofyour step, rearrange and group similar terms using parentheses.(a) 3x2y2 + 5xy + 6xy − x2y2 (b) −4ab5 + 6a + 6ab5 − 2a

This is not difficult in and of itself. It’s a matter of identifying simi-lar terms having, possibly different numerical coefficients, and addingthose coefficients together. Another wrinkle in this process is the prob-lem of subtraction of terms. This is taken up in the next paragraph.


• Adding/Subtracting Grouped ExpressionsThere are a couple of basic rule for combining grouped expressions.

Add/Subtract Groups:

a + (b + c) = a + b + c

a − (b + c) = a − b − c(10)

Procedure. The basic procedure for adding or subtracting algebraicexpressions is

1. remove the grouping parentheses using equations (10);2. rearrange and regroup similar terms, if any;3. combine similar terms by adding numerical coefficients.

Example 3.3. Combine each of the following.(a) (4x3 + 3xy − 2x2) + (5x3 + 9xy − 3x2)(b) (8ab − 2a − 4b) − (−4ab + a − 3b)


Exercise 3.8. Combine each of the following expression. The princi-ple tools to be used: The standard procedure for combining groupedexpressions, equations (10) and equation (2).(a) (4x − 6y) + (7y − 2x)(b) (5xy2 − 3x2y + 2x) − (3x2y − 7xy2)(c) (a − b + c) − (3a − 4b + c) − (4a + b).(d) 4x − [ 6x − (12x + 3) ].

Exercise 3.9. Part (d) of Exercise 3.8 was an example of nestedparentheses. In case you missed that one, here are a couple more topractice on. Passing grade is 100%.(a) 4ab − [ 5ab + (4 − 3ab) ](b) x − (4x + 2y) − [ 6x − 2y − (4x − 4y) ](c) x − (4x + 2y) − [−(6x − 2y) − (4x − 4y) ]

You have reached the end of Lesson 3. In the next lesson, we’ll takeup multiplication, expanding, and combining fractions by getting a


common denominator. Many of the topics of this lesson will be revis-ited in the next. See you in Lesson 4!


3.1. Solutins: Carefully remove each layer of parentheses.(a) Putting x = 3 in x[ 1 + (x + 1)1/2 ]2 becomes

3[ 1 + (1 + 3)1/2 ]2 = 3[ 1 + 41/2 ]2

= 3[ 1 + 2 ]2 = 3[ 3 ]2

= 3(9) = 27

(b) Putting x = −2 in (2x2)3 becomes

( 2(−2)2 )3 = ( 2(4) )3 = 83 = 512

(c) Putting x = 0 in (x + 1){ 1 + [ 1 + (x + 1)2 ]2 }2 becomes

(0 + 1){ 1 + [ 1 + (0 + 1)2 ]2 }2 = { 1 + [ 1 + 1 ]2 }2 = { 1 + 22 }2

= { 1 + 4 }2 = { 5 }2 = 25

Did you get them all? Were you slow and methodical? Exercise 3.1.


3.2. Solution:

At x = 1, 1/2 + x becomes 1/2 + 1 = 3/2 =32.

At x = 1, 1/(2 + x) becomes 1/(2 + 1) = 1/3 =13.

The values of these two expressions are different at x = 1; therefore,I conclude that 1/2+x and 1/(2+x) are not equivalent. I resolve notto equate them in the future! Exercise 3.2.


3.3. Solutions:(a) (−x)5 = ( (−1)x )5 = (−1)5x5 = −x5.(b) (−x)2k = (−1)2kx2k =

((−1)2

)kx2k = 1kx2k = x2k.

(c)[−x3y2

z4

]3

=(−1)3(x3)3(y2)3

(z4)3=

−x9y6

z12 .

Exercise Notes: Problem (a) contains a few more details, problems(b) and (c) accelerates the process of simplifying: We go immediatelyfrom x2k, for example, to (−1)2kx2k.

Exercise 3.3.


3.4. Solutions:

(a) −(−3x + 5y) = 3x − 5y.(b) −(9 − (−2x)3) = −(9 − (−8x3)) = −(9 + 8x3) = −9 − 8x3.(c) −(xy − (3xy − 4)) = −(xy − 3xy +4) = −(−2xy +4) = 2xy − 4.(d) Consider the following calculation:

[1 − (−1)3] −[( 1

2

)3 − (− 12

)3]= [1 − (−1)] −

[18 − (− 1

8

)]= [1 + 1] −

[18 + 1

8

]= 2 − 1

4

= 74 .

If you carefully apply the rules you will surely not error. It’sonly when you become the master of algebra can you acceleratethe simplification process. Take it slow and methodical.

Exercise 3.4.


3.5. Solutions:

(a)[xy + 1

x + y

]−1=

x + y

xy + 1.

(b)[ (x + y)4

x1/2y3/2

]−4=

[x1/2y3/2

(x + y)4]4

=x4/2y12/2

(x + y)16=

x2y6

(x + y)16.

(c)[ ab−1

(a − b)−4

]−2=

[ (a − b)−4

ab−1

]2=

(a − b)−8

a2b−2 =b2

a2(a − b)8.

(d) (x3y4)−2 =1

(x3y4)2=

1x6y8 .

(e)1

x2/y4 =y4

x2 .

(f)[ −x2

y3/z4

]3=

[−x2z4

y3

]3=

(−x2)3(z4)3

(y3)3=

−x6z12

y9 .

Exercise 3.5.


3.6. Solutions: These are simple!(a) 5xyz2 + 6xyz2 − 8xyz2 = 3xyz2.(b) 6x

√x − 3

2x√

x = 92x

√x.

Exercise 3.6.


3.7. Solutions: ı(a) Combine 3x2y2 +5xy+6xy −x2y2 using paren-theses to group similar terms:

3x2y2 + 5xy + 6xy − x2y2 = (3x2y2 − x2y2) + (5xy + 6xy)

= 2x2y2 + 11xy.

ı(b) Combine −4ab5 + 6a + 6ab5 − 2a.−4ab5 + 6a + 6ab5 − 2a = (−4ab5 + 6ab5) + (6a − 2a)

= 2ab5 + 4a

Exercise 3.7.


3.8. Solutions:(a) (4x − 6y) + (7y − 2x) = (4x − 2x) + (−6y + 7y) = 2x + y

(b) Combine (5xy2 − 3x2y + 2x) − (3x2y − 7xy2)

(5xy2 − 3x2y + 2x) − (3x2y − 7xy2)

= 5xy2 − 3x2y + 2x − 3x2y + 7xy2

= (5xy2 + 7xy2) + (−3x2y − 3x2y) + 2x

= 12xy2 − 6x2y + 2x

(c) Combine (a − b + c) − (3a − 4b + c) − (4a + b).

(a − b + c) − (3a − 4b + c) − (4a + b)= a − b + c − 3a + 4b − c − 4a − b

= (a − 3a − 4a) + (−b + 4b − b) + (c − c)

= −6a + 2b


(d) Combine 4x − [ 6x − (12x + 3) ].

4x − [ 6x − (12x + 3) ]

= 4x − 6x + (12x + 3)= 4x − 6x + 12x + 3

= (4x − 6x + 12x) + 3

= 10x + 3

Comments: The above solutions have more detail than is usually in-clude when a master student of algebra performs. If you are not amaster of algebra yet, take is slow and methodically. After a few prob-lems, you can get the “feel” of the operations and you can accelerateyou solutions. But don’t do it until you are a master ! Exercise 3.8.


3.9. Solutions: In each case, you should have carefully removedparentheses.(a) 4ab − [ 5ab + (4 − 3ab) ].

4ab − [ 5ab + (4 − 3ab) ]

= 4ab − 5ab − (4 − 3ab)= 4ab − 5ab − 4 + 3ab

= 2ab − 4 � accelerate a bit!

(b) x − (4x + 2y) − [ 6x − 2y − (4x − 4y) ].

x − (4x + 2y) − [ 6x − 2y − (4x − 4y) ]

= x − 4x − 2y − 6x + 2y + (4x − 4y)= x − 4x − 2y − 6x + 2y + 4x − 4y

= (x − 4x − 6x + 4x) + (−2y + 2y − 4y)

= −5x − 4y


(c) x − (4x + 2y) − [−(6x − 2y) − (4x − 4y) ].

x − (4x + 2y) − [−(6x − 2y) − (4x − 4y) ]

= x − 4x − 2y + (6x − 2y) + (4x − 4y)= x − 4x − 2y + 6x − 2y + 4x − 4y

= (x − 4x + 6x + 4x) + (−2y − 2y − 4y)

= 7x − 8y

Did you get them all right? It is important not to error! Concentrate!Concentrate! And . . . oh yes, concentrate! Focus your mind on theproblem at hand! Exercise 3.9.


3.1. Two methods are demonstrated.

Method 1: Working from the outer most to the inner most.

−[x − (−4x + y)] = −x + (−4x + y) � Negate a sum

= −x − 4x + y

= −5x + y

Method 2: Working from the inner most to the outer most.

−[x − (−4x + y)] = −[x + 4x − y] � Negate a sum

= −[5x − y]= −5x + y

Example 3.1.


3.2. Solutions

(a)1

x3/y2 =y2

x3 .

(b)1

x/(x + y)=

x + y

x. Notice that I have used parentheses to

clearly define the expression under consideration.

(c)1

x/(x + y)−3 =(x + y)−3

x=

1x(x + y)3

. In the last step, equa-

tion (5) was used.Example 3.2.


3.3. The principle tools are equations (10) equation (2). We followthe standard procedure:

Solution to (a):(4x3 + 3xy − 2x2) + (5x3 + 9xy − 3x2)

= 4x3 + 3xy − 2x2 + 5x3 + 9xy − 3x2 (S-1)

= (4x3 + 5x3) + (3xy + 9xy) + (−2x2 − 3x2) (S-2)

= 9x3 + 12xy − 5x2 (S-3)

Comments: Step (1) of the procedure was followed in (S-1); Step (2)was followed in (S-2), and (S-3) represents Step (3)


Solution to (b): The solution to this part follows the same pattern aspart (a), so there will be fewer annotations.

(8ab − 2a − 4b) − (−4ab + a − 3b)

= 8ab − 2a − 4b + 4ab − a + 3b (S-4)

= (8ab + 4ab) + (−2a − a) + (−4b + 3b)

= 12ab − 3a − b

Comments: In line (S-4), we removed the parentheses by followingthe standard rules. Here is more detail. The problem is basically ofnegation:

−(−4ab + a − 3b) = −(−4ab) − a − (−3b) � by (10) or (4)

= 4ab − a + 3b � from (2)

The last expression was then put back into equation (S-4) and thecalculation continued. Example 3.3.

Important Points


Right on! Let’s have some discussion in case you missed the first time.

Question:Is x2 a common factor to all terms of the expression x3y −4x6 + 3x7/2y3?

The answer is “Yes.” Indeed, do a rewrite as follows.

x3y − 4x6 + 3x7/2y3 = (x2)(xy)︸︷︷︸first term

− (x2)(4x4)︸︷︷︸second term

+(x2)(3x3/2y3)︸︷︷︸third term

When written in this manner, you can now see that each term has thefactor x2 in common. It is a very important skill to have the ability to“spot” common factors of terms of an expression. Important Point


Yes, he is forced to write −32 to describe the operations of “minus thenumber three squared. But now he is in the sticky situation of usingthe same notation for two entirely different sequences of operations.

In the student’s mind 1: −32 means, in some situations, (−3)(−3) = 9and . . .

In the student’s mind 2: −32 means, in some situations, −3 · 3 = −9.

Same notation, different results. Bad situation, indeed.Important Point


The correct answer is (b): The expression x + x2 evaluates to 0 whenwe put x = −1; indeed,

(−1) + (−1)2 = (−1) + 1 = 0.

If your original answer was −2 you probably made the fundamentalalgebraic blunder just discussed. This calculation is wrong!

Wrong! =⇒ − 1 − 12 = −1 − 1 = 2 ⇐= Wrong!

If you did this, don’t do it again! DPS Important Point

��mptiimenu


Lesson 4: Basic Algebra, Part II

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 4: Basic Algebra, Part II

Table of Contents4. Basic Algebra, Part II4.1. The Distributive Law4.2. On the Cancellation of Factors4.3. How to Multiply Fractions

• Rationalizing the Denominator4.4. How to Add Fractions

• How to get a Least Common Denominator • Separating aFraction

4. Basic Algebra, Part II

4.1. The Distributive Law

In the process of manipulating algebraic expressions, we often findourselves multiplying expression wherein one or both factors are sums.The Distributive Law tells us how to properly handle productsinvolving sums.

The Distributive Law:Let a, b, and c represent numbers. Then

a(b + c) = ab + ac (1)

In the age of the computer, we might say that the Distributive Lawis a statement of how multiplication interfaces with addition. It is oneof the major tools used in algebra.

Section 4: Basic Algebra, Part II

Using the formulaa(b + c) = ab + ac (2)

from left-to-right might be referred to as expanding or distributingthrough and using it from right-to-left is referred to as factoring.

Reading formula (2) from left-to-right, we can see that when multi-plying one factor, a, by the sum of two quantities, b and c, we simplymultiply a by each term of the sum, and add the result. This is referredto as distributing a through the sum.

Equation (2) is a property of our arithmetic system. Is it true that

4(2 + 5) = 4 · 2 + 4 · 5?Of course, both sides evaluate to 28. This property of our arithmeticsystem—rarely used when dealing with only numerical values—is ex-tremely useful with dealing with symbolic quantities.


Time Out! In the previous paragraph I made the statement that thedistributive law is “rarely used when dealing with only numerical val-ues.” Is that a true statement? E-mail me if you have an exampleof the distributive law in everyday life! D. P. Story.

Illustration 1. Expand each using (1).(a) x(y + z) = xy + xz and x(y − z) = xy − xz. Of course we can

expand or distribute through even when there is subtractionpresent.

(b) x3(x5 + 4x3 + 2) = x8 + 4x6 + 2x3, where we have distributedthrough x3 and combined the powers of x using the first Lawof Exponents.

(c) wt(w3t − 5w−1t2) = w4t2 − 5t3. Once again we combine similarpowers by adding exponents per the first Law of Exponents.

(d) The a, b, and c in (2) may be complex algebraic expressionsthemselves:

(x + y)(a + b) = (x + y)a + (x + y)b.

[email protected]


Here, we have distributed the factor (x + y) through the sum;alternately, we can distribute the (a + b) through the sum:

(x + y)(a + b) = (a + b)(x + y) = (a + b)x + (a + b)y.

These are just simple applications of the Distributive Law.(e) Continuation. Usually, when we have an expression of the form

(x+y)(a+b), the task is to “multiply it out.” Let’s continue thatprocess—all we do is to apply the Distributive Law again:

(x + y)(a + b) = (x + y)a + (x + y)b � Dist. Law

= (xa + ya) + (xb + yb) � Dist. Law

= xa + ya + xb + yb

The point is that all you do is carefully apply the DistributiveLaw when ever you are multiplying out algebraic expressions.

Illustration Notes: Expanding expressions of the type discussed inexamples (d) and (e) above will be taken up in Lesson 5.


Exercise 4.1. Expand each of the following expressions completely,combining similar exponentials using the Law of Exponents wher-ever appropriate. Slowly and methodically work out your solutionsfirst using good notation and techniques.(a) ab(c + d) (b) w−1(4w − 3w2) (c) x3(

√x + x3/2)

(d)√

w(√

w + 3) (e) s3(5s−4 − 3s2 ) (f) (st)2(s−1 + t−1)

Exercise 4.2. Simplify

s−1/2t1/2[

1(st)1/2 + (st)1/2

]

Using the Distributive Law and the Law of Exponents, eliminating allnegative exponents.

The formula a(b+ c) = ab+ ac when read from right-to-left is used intwo ways: combining similar terms and simple factoring

We can reason as follows. The expression ab+ac has two terms. Noticethat each term has a factor of a. (Here we say that a is a common


factor to all the terms.) We can factor out this common factor toobtain,

ab + ac = a(b + c)

In the illustration below, we use this formula in the following form:

ba + ca = (b + c)a. (3)

Same formula, I’ve just written the common factor on the right, andfactored it out to the right.

Illustration 2. Combine similar terms using (3).(a) 4x + 6x = (4 + 6)x = 10x. The first equality comes from the

Distributive Law, read from right-to-left; in particular equa-tion (3).

(b) 5w3 − 2w3 = (5 − 2)w3 = 3w3, by equation (3).(c) 4x3/2 + 6x3/2 − 8x3/2 = 2x3/2.(d) After reviewing the two previous examples, you should realize

that there is nothing difficult about the operation of adding simi-lar terms together. The process of combining can be accelerated:4st + 12st = 16st.


(e)√

x + 1 − 12

√x + 1 = 1

2

√x + 1.

(f)5

w2 + 1+

4w2 + 1

=9

w2 + 1.

Exercise 4.3. Combine similar terms of each of the following.(a) 4w − 12w (b) 23p2q3 − 6p2q3 (c) 1

2 t3 + 32 t3

(d)3s2 − 6

s2 (e)4w√

s− 3w√

s(f) 4xy − 5st

The Distributive Law can be used in a more general way than wasjust illustrated. Reading the Distributive Law from right-to-leftcan be thought of as factoring out all factors common to all terms ofan expression.

Illustration 3. Using (3), factor out any factors common to all theterms. The Law of Exponents is used throughout these examples.(a) xy3 − xy2 = xy2(y − 1). The validity of this step can be verified

by distributing xy2 back through y − 1.(b) 4s2t3 − 8st = 4st(st2 − 2). Again, multiply back through to

verify.


(c) 2x3/2 + 4x5/2 = 2x3/2(1 + 2x). Here, we have factored x5/2 asx5/2 = x3/2x, by the Law of Exponents, then factored out acommon factor of x3/2.

(d) Here’s an example with more detail thrown in.

4x3y6z − 12x4y7 = (4x3y6)z − (4x3y6)(3xy) � Law Exp. #1

= (4x3y6)[z − (3xy)] � Distrib. Law

= 4x3y6(z − 3xy)

(e)st3

w2 − st

w2 =st

w2 (t2 − 1) =st(t2 − 1)

w2 .

(f)x

w− y

w=

1w

(x − y) =x − y

w.

Illustration Notes: Examples (e) and (f) are tantamount to addingtogether two fractions having the same denominator.Operational Tips. Here are two observations that may aid you infactoring out using the Distributive Law.


� Tip 1: When factoring out common factors involving exponents,factor out the powers having the smallest exponent:

x4y3 + x2y5 = x2y3(x2 + y2).

Here, I factored out x2 because it was the smallest power of xand I factored out y3 because it was the smallest power of y.

� Tip 2: Notice that when factoring out the lowest common powers

x8y5 + x3y7 = x3y5(x8−3y5−5 + x3−3y7−5) = x3y5(x5 + y2)

the calculation of the powers of the remaining factors is madeby subtracting off the factored power, as illustrated above.

Review Illustration 3 and observe these Operational Tips in ac-tion. Then apply them to the following . . .

Exercise 4.4. Factor out all factors that are common to all the termsof the algebraic expressions below.(a) 3w3v4 + 9w2v2 (b) x2√y + x3y3/2 (c) ab2c3 + a3b2c

(d)xy

5− xy

10(e) (x + 1)

√x − x

√x (f)

2p8q4

w2 +4p8

w


The Operational Tips are still valid for negative powers as well. Deal-ing with negative powers is when students find themselves most atrisk for errors. The rules are simple: Factor out the smallest power(Tip 1), by subtracting it form the others (Tip 2).

Note: It is subtracting a negative number that causes the problems:Remember,

2 −(−1

2

)= 2 +

12

=52. (4)

a “minus times a minus is a plus.”

Illustration 4. Factor out any common factors present.(a) 4x2 −2x−1/2 = 2x−1/2(2x5/2 −1). The details of the calculation

of the exponent is given in equation (4).(b) 3x1/2y−2/3 − 12x−3/2y−5/3 = 3x−3/2y−5/3(x2y − 4). Verify that

the exponents in x2 and y are the correct ones as calculated byTip 2.

Let’s have one more set of exercises to illustrate this last point.


Exercise 4.5. While taking the advice of the Operational Tips, fac-tor out all common factors.(a) x8y−2 − x−3y−4 (b) 6a3/2 − 9a−1/2

(c) 12w−3/2s−12/5 + 8w−1/2s−2/5

4.2. On the Cancellation of Factors

What do I mean by that? I refer to the rather profound resulta

a= 1 a �= 0.

Most often this fact is used in the following context:

ab

ac=

b

c

and students, operationally, give it this action,

ab

ac=

/ab

/ac=

b

c,

though I would never do that myself.


Let’s elevate this to the status of a shadow box.

Cancellation Law

ab

ac=

b

ca �= 0, c �= 0 (5)

The so-called Cancellation Law has already been visited by theselessons. When we discussed the Law of Exponents in Lesson 2,we advertised a version of Law #1 as a Cancellation Law. (How’sthat for fancy hypertexting?)

We have already seen that

an

am= an−m.

Our new Cancellation Law is just the case of n = m = 1. Conse-quently, very little time will be devoted to cancellation. What we willdo is to concentrate on cancellation after factoring.


The Cancellation Law to can be verbalized as follows:ab

ac=

b

c

states that if a is a factor of the numerator and the denominator, thenthe a can be cancelled from the expression.

Here are a few examples.

Illustration 5. Simplify each of the following.

(a)xy − xy2

xy=

(xy)(1 − y)xy

= 1 − y, where we have cancelled the

factor xy from the numerator and denominator.

(b)√

x + x√

x

x√

x=

√x(1 + x)x√

x=

1 + x

x. Here’s an important point.

Some students will cancel the x’s as well. This is illegal. Thereason for this is that x is not a common factor of both thenumerator and denominator.

(c)x2y3z5 − 4x3y2z5

xy4z5 =x2y2z5(y − 4x)

xy4z5 =x(y − 4x)

y2 .


Illustration Notes: You must train your eyes to see common factors,automatically factor them out, and cancel if appropriate. This is avery important skill.Exercise 4.6. Simplify each of the following fractions by factoringout common factors and cancelling.

(a)a + ab

a(b)

x2y4z9

x3y5z3 (c)x−3y−2

x4y−5

(d)3x3 + 9x2

3x3 (e)4x5/2 − x7/2

x2 (f)y3√x − y2√x

xy2√

x

Don’t forget, negative powers can be factored out as well.

Exercise 4.7. Simplify each of the following fractions by factoringout common factors and cancelling. After factoring and cancelling,remove all negative exponents.

(a)x−3y−2 − x2y−5

x4y−5 (b)4x−5/2 − x7/2

x1/2 (c)x2

4x3 + 2x−3


4.3. How to Multiply Fractions

Multiplying fractions together est du gateau.1 We have been multi-plying fractions throughout these lessons. For any a, b, c and d, wehave

a

b

c

d=

ac

bd(6)

When multiplying two fractions together, just multiply the numera-tors and divide by the product of the denominators.

Illustration 6.

(a)12

x

x + y=

x

2(x + y).

(b)x3y

(x + y)z2

(x + y)=

x3yz2

(x + y)2.

1Pardon my French. Translation: Multiplying fractions together is a (piece) ofcake.


(c) Any number or symbol can be thought of as a fraction; for ex-

ample, 4 =41

and x =x

1. In light of this observation, consider

the following example:

4 · x

2x − 1=

41

· x

2x − 1=

4x2x − 1

.

This simplification is usually accelerated; we jump from the firstto the last expression in one step.

(d) The generalization of the previous point would be, for any a,and b �= 0,

a

b= a · 1

b. (7)

(e) (x+1) · y

z· x

(x + 1)=

(x + 1)yx

z(x + 1)=

xy

z, where we have cancelled

the common factor (x + 1).

Time for you to chime in. Try the following . . .


Exercise 4.8. Combine all fractions, simplify as appropriate.

(a)4x2

5y3

x4y2

x2y(b) (3x − 2)2

4x(x + 1)3

(c)x

y

[y

z+

12

]

(d)[ x4

(x + 3)· x

2

]−1(e)

[x(x

y

)−1]−3

Before leaving this sections, let’s discuss ratios of fractions; that is,expressions of the form

a/b

c/d=

a

bc

d

.

The question is how can this be simplified? This question is easilyanswered using the techniques of this lesson. Observe that

a

bc

d

=a

b· 1

c

d︸︷︷︸from (7)

=a

b· d

c

︸︷︷︸from (8)

=ad

bc

︸︷︷︸by (6)


Thus, the governing equation for simplifying multifractions is

a/b

c/d=

a

bc

d

=ad

bc(8)

Exercise 4.9. Simplify each of the following using (8).

(a)x/(x + y)

x2/(x + y)2(b)

x3y2/s9t2

x4y8/s10 (c)x4

x3/4(d)

xy/(x + y)(x + y)

• Rationalizing the DenominatorThe traditional rules for simplifying algebraic expressions frown onhaving radicals in the denominator. There is a simple device for ap-peasing the rules in this case it’s called rationalizing the denominator :For example,

1√2

=1√2

·√

2√2

=√

22


That is, we multiply the numerator and denominator by the appro-priately chosen radical for the purpose of ridding ourselves of radicalsin the denominator.

Illustration 7. Rationalize the denominator in each of the follow-ing.

(a)4√3

=4√3

·√

3√3

=4√

33

.

(b) Rationalizing is not restricted to numerical expressions.

21y√x

=21y√

x·√

x√x

=21y

√x

x.

(c)4xy√x2 + 1

=4xy√x2 + 1

·√

x2 + 1√x2 + 1

=4xy

√x2 + 1

x2 + 1.

(d) This trick is not restricted to the square root either:

53√

2=

521/3 · 22/3

22/3 =5(22/3)

2=

5 3√

42


Exercise 4.10. Rationalize the denominator of each of the following.

(a)x√7

(b)3x2

√x + 1

(c)xy4√

3

4.4. How to Add Fractions

Adding fractions is always a sore point among students, but it reallyisn’t that difficult. Here is the guiding principle: When adding twofractions with the same denominator, simply add the numerators anddivide by the common denominator. In symbols:

Addition of Fractions:a

c+

b

c=

a + b

c(9)

This is the simplest situation. Let’s look at some examples.

Illustration 8. In each case, we already have a common denomina-tor; consequently, adding or subtracting fractions is trivial pursuit.


(a)39

+59

=89. The two fractions have a common denominator, so

we add the numerators. This is a skill we have been working allour lives to perfect.

(b)3x + 4

5+

6x − 55

=(3x + 4) + (6x − 5)

5=

9x − 15

.

(c)4s

s + t− 5s

s + t=

4s − 5ss + t

=−s

s + t= − s

s + t.

The next question is, how do you add fractions when don’t have acommon denominator? The answer is that you create a common de-nominator using a simple device. Next up is an enumeration of thesteps to add to fractions together having different denominators.


Problem. Addx

s+

y

t. (Actually, combine would be a better term—

they are already added!)� Step 1: Multiply the first fraction by t/t and the second fraction

by s/s to obtainx

s+

y

t=

x

s

t

t+

y

t

s

s.

� Step 2: Combine fractions for each term.x

s+

y

t=

xt

st+

ys

ts

They now have a common denominator.� Step 3: Add the fractions together.

x

s+

y

t=

xt + ys

st

The idea is to multiply the numerator and denominator by the samequantity in such a way that, at the end of the process, all fractionshave the same denominator.


Example 4.1. Combine each of the following. The common denom-inator is obtained by taking the product of the denominators of eachterm.

(a) 6x − 3x

(b)1√x

− √x (c)

x2

x + y− x

y(d)

2x

− 3ya

+z

2b

Here are a few exercises for you to practice on.

Exercise 4.11. Combine each expression.

(a)2a

+a

2(b)

3x2x − y

+y

x(c)

12y

− x

3+

y

(1 + x)

• How to get a Least Common DenominatorWhen combining the sum of algebraic expressions it is not alwayswise or efficient to obtain a common denominator by simply multi-plying the denominators of each term together. (See the techniquesillustrated in the previous paragraph on How to Add Fractions.) Mostof the time we want to obtain the least common denominator ; thatis, a denominator that is a simple as possible.


Illustration 9. Contrast the following two methods of combining the

expression1x

+2x2 +

3x3 .

(a) Get a common denominator by multiplying the denominatorstogether: x · x2 · x3 = x6. Then

1x

+2x2 +

3x3 =

1x

x5

x5 +2x2

x4

x4 +3x3

x3

x3

=x5

x6 +2x4

x6 +3x3

x6 =x5 + 2x4 + 3x3

x6(10)

(b) An alternate choice of a common denominator is x3. In this casewe get

1x

+2x2 +

3x3 =

1x

x2

x2 +2x2

x

x+

3x3

=x2

x2 +2xx3 +

3x3 =

x2 + 2x + 3x3

(11)


Illustration Notes: Both equations (10) and (11) are valid, but givedramatically different results. Obviously, (10) is unduly complex; infact, x3 is a common factor to both numerator and denominatorin (10) and should be cancelled.

The reason for the disparity between these two answers is thechoice is the denominator. It was a bad choice to use x6. The choiceof x3 is referred to as the least common denominator.

Finding the Least Common Denominator1. Factor each denominator completely.2. Write down a list of all factors.3. Among all factors having the same base, remove all but

the one having the highest exponent.4. Multiply all the factors together left after step 3 to

obtain the least common denominator.

Next is an important example that explains in detail how to use thisLCD algorithm. You should read and study this example.


Example 4.2. Combine14x

− 4x3y

+3x2y4 .

If you have a history of not being able to successfully compute theLCD of an algebraic expression and carry out the process of addingfractions together, go slowly through out the rest of this section. Besure you understand the process.

The LCD Algorithm works equally well with radicals and fractionalexponents.

Example 4.3. Combine the expression:2√

x

5y2 +3

2x1/2y3/2 .

Let’s test your understanding of the algorithm for obtaining the leastcommon denominator. Below is a little quiz for you to take. Considerthe questions carefully before responding. Follow the steps of the LCDAlgorithm.

Quiz. Determine the LCD for each of the following.

1.3x

2y2z3 − 2xy3z2 .


(a) LCD = 2xy5z5 (b) LCD = 2y3z3

(c) LCD = 2xy3z3 (d) LCD = 2xy3z5

2.x + y

3x3/2y2 − x2 + y2

6xy4 .

(a) LCD = 18x3/2y4 (b) LCD = 6x3/2y4

(c) LCD = 18xy4 (d) LCD = 6xy4

End Quiz.

Now that you have seen a couple of examples in detail, you try. Youhave to be very simple minded ! Apply the LCD Algorithm and usegood algebraic techniques. Don’t skip steps; do it once, right the firsttime!

Exercise 4.12. Solve each of the following. I recommend you use theLCD Algorithm and the procedure outlined in Example 4.2 to obtaina least common denominator before combining fractions. Radicals canbe converted to exponential notation. Negative exponents should beeliminated first.


(a)4

3x2y3 − 54x3y

(b)3a4b4 +

a + b

9ab

(c)x2y

2(x + y)− y

4x2(x + y)(d)

1xy5 − 2

x3y4 +3

x5y3 − 4x7y2

• Separating a FractionWhen we obtain a common denominator, we read and use equation (9)from left-to-right. The fundamental equation for adding fractions,equation (9), can also be read from right-to-left. This is a processof separating a fraction.

The equationa + b

c=

a

c+

b

c(12)

states that if we have a quotient and the numerator consists of thesum (or difference) of several terms, then the fraction can be brokenup or separated according to (12).

This technique can be used to simplify an algebraic expression; moreimportantly, within the context of certain (Calculus) problems, using


this separation of fractions technique can advance your work towarda successful conclusion.

Illustration 10. Separate each of the following and simplify.

(a)4x2 + 3

2x2 =4x2

2x2 +3

2x2 = 2 +3

2x2 = 2 +3

2x2 .

(b) Here’s a common situation in which separation is useful.

4x4 − 3x3 + 2x2 − x + 1x

=4x4

x− 3x3

x+

2x2

x− x

x+

1x

= 4x3 − 3x2 + 2x − 1 +1x

.

(c)4y − 1y + 1

=4y

y + 1− 1

y + 1.

Here are a couple of problems for you to consider.

Exercise 4.13. Separate each of the following and simplify.


(a)12x3 − 6x2 + 3

3x2 (b)x + 1x2 + 1

(c)2(x + 1)2 + 6x

(x + 1)

Common Error. An error students make in the heat of battle (a test)is the following one:

NOT TRUE! =⇒ a

b + c=

a

b+

a

c⇐= NOT TRUE!

This is called an algebraic blunder! Don’t do this!

We’ve now finished Lesson 4. If you’ve stuck with me, congratu-lations! Follow me now to Lesson 5 to review multiplication anddivision of algebraic expressions.


4.1. Solutions: Throughout we use the Law of Exponents withoutcomment.(a) ab(c + d) = abc + abd.(b) w−1(4w − 3w2) = 4 − 3w.(c) x3(

√x + x3/2) = x3(x1/2 + x3/2) = x7/2 + x9/2.

(d)√

w(√

w + 3) = w + 3√

w.

(e) s3(5s−4 − 3s2 ) = 5s−1 − 3s3

s2 =5s

− 3s.

(f) Here is a bit more detail for this one

(st)2(s−1 + t−1) = s2t2(s−1 + t−1) � Law Exp. #2

= s2t2s−1 + s2t2t−1� Distrib. Law.

= s2−1t2 + s2t2−1� Law Exp. #1

= st2 + s2t � arithmetic


A similar documentation can be done for problems (a)–(e); didyou document?

Your answers may not jive exactly with what I have . . . of course, youshould continue to manipulate your answers to get mine.

Exercise 4.1.


4.2. Solution.

s−1/2t1/2[

1(st)1/2 + (st)1/2

]

=s−1/2t1/2

(st)1/2 + s−1/2t1/2(st)1/2� Distrib. Law

=s−1/2t1/2

s1/2t1/2 + s−1/2t1/2s1/2t1/2� Law Exp. #2

=1

s1/2+1/2 + s−1/2+1/2t1/2+1/2� Law Exp. #1

=1s

+ t

We have used the fact that s0 = 1 and t0 = 1. Exercise 4.2.


4.3. Solution:(a) 4w − 12w = −8w.(b) 23p2q3 − 6p2q3 = 17p2q3.(c) 1

2 t3 + 32 t3 = 2t3.

(d)3s2 − 6

s2 = − 3s2 .

(e)4w√

s− 3w√

s= 4

w√s

− 3w√s

= (4 − 3)w√s

=w√s.

(f) 4xy − 5st does not combine any further. : -)Exercise 4.3.


4.4. Solutions:(a) 3w3v4 + 9w2v2 = 3w2v2(wv2 + 3).(b) x2√y + x3y3/2 = x2y1/2 + x3y3/2 = x2y1/2(1 + xy).(c) ab2c3 + a3b2c = ab2c(c2 + a2) = ab2c(a2 + c2).

(d)xy

5− xy

10=

xy

5

[1 − 1

2

]=

xy

512

=xy

10.

(e) (x + 1)√

x − x√

x = [(x + 1) − x]√

x =√

x. Here have factored√x out to the right. I hope that didn’t psychologically disturb

you.

(f)2p8q4

w2 +4p8

w=

2p8

w

[q4

w+ 2

].

Problem (f) can be simplified further, if you allow me to get a commondenominator.

2p8q4

w2 +4p8

w=

2p8

w

[q4

w+ 2

]

=2p8

w

q4 + 2ww

=2p8(q4 + 2w)

w2 .

Exercise 4.4.


4.5. Solutions:(a) x8y−2 − x−3y−4 = x−3y−4(x11y2 − 1).

(b) 6a3/2 − 9a−1/2 = 3a−1/2(2a2 − 3).

(c) 12w−3/2s−12/5 + 8w−1/2s−2/5 = 4w−3/2s−12/5(3 + 2ws2).

Comments: If you did not get these problems correct, be sure youunderstand why! The calculation of the exponents is always the prob-lem. Exercise 4.5.


4.6. Solutions:

(a)a + ab

a=

a(1 + b)a

= 1 + b.

(b)x2y4z9

x3y5z3 =z6

xy.

(c)x−3y−2

x4y−5 =y−2y5

x4x3 =y3

x7 .

(d)3x3 + 9x2

3x3 =3x2(x + 3)

3x3 =x + 3

x.

(e)4x5/2 − x7/2

x2 =x5/2(4 − x)

x2 = x1/2(4 − x) =√

x(4 − x).

(f)y3√x − y2√x

xy2√

x=

y2√x(y − 1)xy2

√x

=y − 1

x.

Cancellation in many cases is carried out by the subtraction of expo-nents. For example, in problem (e), we simplified the fraction

x5/2

x2 = x5/2x−2 = x5/2−2 = x1/2 =√

x.


If you become “expert” at these kinds of simplifications, you can jumpfor the left-most expression to the right-most expression in one “mag-nificent step.” Exercise 4.6.


4.7. Solutions:

(a) Simplifyx−3y−2 − x2y−5

x4y−5 .

x−3y−2 − x2y−5

x4y−5 =x−3y−5(y3 − x5)

x4y−5 � Tip 1 & Tip 2

=x−3(y3 − x5)

x4 � cancel

=y3 − x5

x7 . � Law of Exp. #1

The details of the other two problems are similar. I’ll just presentthe answers. Mimic the solution to (a) to obtain correct solutionsto (b) and (c) as necessary.

(b)4x−5/2 − x7/2

x1/2 =4 − x6

x3

(c)x2

4x3 + 2x−3 =x5

2(2x6 + 1).

Exercise 4.7.


4.8. Solutions:

(a)4x2

5y3

x4y2

x2y=

4x2x4y2

5y3x2y=

4x6y2

5x2y4 =4x4

5y2 .

(b) (3x − 2)24x

(x + 1)3=

4x(3x − 2)2

(x + 1)3

(c)x

y

[y

z+

12

]=

x

y

y

z+

x

y

12

=xy

yz+

x

2y=

x

z+

x

2y.

(d)[ x4

(x + 3)· x

2

]−1=

[ x5

2(x + 3)

]−1=

2(x + 3)x5 .

(e)[x(x

y

)−1]−3=

[x(y

x

)]−3=

[xy

x

]−3= y−3 =

1y3 .

You should fully understand the justification of each step. Knowledgeof the validity of each step will give you confidence in your algebraicabilities. Exercise 4.8.


4.9. Solutions: We simply apply (8).

(a)x/(x + y)

x2/(x + y)2=

x(x + y)2

x2(x + y)=

x + y

x.

(b)x3y2/s9t2

x4y8/s10 =x3y2s10

s9t2x4y8 =s

xy6t2.

(c)x4

x3/4=

4x4

x3 = 4x.

(d)xy/(x + y)

(x + y)=

xy

(x + y)2.

In the case of (c), we have (mentally) writtenx4

x3/4as

x4/1x3/4

and thenapplied (8).

Ditto for part (d). Writexy/(x + y)

(x + y)as

xy/(x + y)(x + y)/1

and apply (8).Exercise 4.9.


4.10. Solutions:

(a)x√7

=√

7x

7.

(b)3x2

√x + 1

=3x2√x + 1

x + 1.

(c)xy4√

3=

xy

31/4 =33/4xy

3=

(33)1/4 xy

3=

4√

27xy

3.

Exercise 4.10.


4.11. Solutions:

(a)2a

+a

2=

2a

22

+a

2a

a=

42a

+a2

2a=

4 + a2

2a(b) We proceed as follow:

3x2x − y

+y

x=

3x2x − y

x

x+

y

x

2x − y

2x − y=

3x2

x(2x − y)+

y(2x − y)x(2x − y)

=3x2 + y(2x − y)

x(2x − y)=

3x2 + 2xy − y2

x(2x − y)

(c) Let’s use standard techniques.

12y

− x

3+

y

(1 + x)=

12y

3(x + 1)3(x + 1)

− x

32y(x + 1)2y(x + 1)

+y

(1 + x)6y6y

=3(x + 1) − 2xy(x + 1) + 6y2

6y(x + 1)

=−2x2y − 2xy + 6y2 + 3x + 3

6y(x + 1)Exercise 4.11.


4.12. Solutions: Fill in the details please!

(a)4

3x2y3 − 54x3y

=16x − 15y2

12x3y3

(b)3a4b4 +

a + b

9ab=

(3a)(9a) + (a + b)(4b3)36ab4 =

27a2 + 4ab3 + 4b4

36ab4

(c) Here the LCD = 4x2(x + y). Thus,

x2y

2(x + y)− y

4x2(x + y)=

(x2y)(2x2) − y

4x2(x + y)

=2x4y − y

4x2(x + y)

(d) The LCD = x7y5. Thus,

1xy5 − 2

x3y4 +3

x5y3 − 4x7y2 =

x6 − 2x4y + 3x2y2 − 4y3

x7y5

Exercise 4.12.


4.13. Solutions:

(a)12x3 − 6x2 + 3

3x2 =12x3

3x2 − 6x2

3x2 +3

3x2 = 4x − 2 +1x2

(b)x + 1x2 + 1

=x

x2 + 1+

1x2 + 1

(c)2(x + 1)2 + 6x

(x + 1)=

2(x + 1)2

(x + 1)+

6x(x + 1)

= 2(x + 1) +6x

(x + 1)

Exercise 4.13.


4.1. Solutions:

(a) Combine 6x − 3x.

6x − 3x

= (6x)x

x− 3

x=

6x2

x− 3

x� get common denom.

=6x2 − 3

x=

3(2x2 − 1)x

� combine

(b) Combine1√x

− √x.

1√x

− √x =

1√x

− √x

√x√x

� get common denom.

=1√x

− x√x

=1 − x√

x� combine


(c) Combinex2

x + y− x

y.

x2

x + y− x

y=

x2

x + y

y

y− x

y

(x + y)(x + y)

� get common denom.

=x2y

(x + y)y− x(x + y)

y(x + y)� common denom. attained!

=x2y − x(x + y)

(x + y)y� combine

=x2y − x2 − xy

(x + y)y

(d) Combine2x

− 3ya

+z

2b.

2x

− 3ya

+z

2b=

2x

2ab

2ab− 3y

a

2xb

2xb+

z

2bax

ax� get common denom.

=4ab − 6bxy + axz

2abx


Example Notes: The common denominator is obtained by multiplyingall the denominators together. Once the common denominator hasbeen identified, we multiply the numerator and denominator of eachterm by 1 in a form that will produce the common denominator. (Thisshould be clear!)

Examples (a)–(c) demonstrate the procedure for combining twofractions. In example (d), the same procedure is used for obtaining acommon denominator for the sum of three terms.

This is the same procedure that you have been using for yearsto combine numerical fractions: 1/3 − 1/2 = 2/6 − 3/6 = −1/6.

This is a very important and basic skill. Combining is simplic-ity itself as long as you understand the underlying reasoning—thatof multiplying each term by an appropriately chosen version of 1.Whenever you get confused and/or disoriented about what to do, theproblem reduced down to multiplying each term by 1!

You can always obtain a common denominator by multiplyingall the denominators together. Usually though, we try to obtain theleast common denominator ! Details to follow. Example 4.1.


4.2. To find the LCD of14x

− 4x3y

+3x2y4

I’ll follow my own advice.

Step 1: Factor each denominator completely. In this example,this step is easy, though Step 1 has the potential of being the mostdifficult step.

First term has denominator: 4x Factors: 22, xSecond term has denominator: x3y Factors: x3, yThird term has denominator: 2y4 Factors: 2, y4

Step 2: Write down all factors. The factors are

22, x, x3, y, 2, y4

Step 3: Remove lowest powers. Let’s rearrange the previous dis-play:

2, 22,︸︷︷︸same base

x, x3,︸︷︷︸same base

y, y4︸︷︷︸same base


Within each group, we eliminate all but the one with the highestexponent to obtain

22, x3, y4

Step 4: Multiply all the factors together left after step 3 toobtain the least common denominator. According to my LCD“algorithm,” the least common denominator is

LCD = 22x3y4 = 4x3y4.

Now that we have jumped through hoops to get the LCD, we stillhave the job of combining the expression.


Now combine!14x

− 4x3y

+3x2y4 =

14x

[x2y4

x2y4

]− 4

x3y

[4y3

4y3

]+

3x2y4

[2x3

2x3

]

=x2y4

4x3y4 − 16y3

4x3y4 +6x4

4x3y4

=x2y4 − 16y3 + 6x4

4x3y4

Presentation of Answer :

14x

− 4x3y

+3x2y4 =

x2y4 − 16y3 + 6x4

4x3y4

That was easy!

Once you understand and understand the LCD Algorithm, the processof getting the least common denominator will be . . . second nature.

Example 4.2.


4.3. Let’s delineate the details one more time! Our given expressionis

2√

x

5y2 +3

2x1/2y3/2 .

Step 1: Factor each denominator completely.First term has denominator: 5y2 Factors: 5, y2

Second term has denominator: 2x1/2y3/2 Factors: 2, x1/2, y3/2

Step 2: Write down all factors. The factors are

5, y2, 2, x1/2, y3/2

Step 3: Remove lowest powers. Let’s rearrange the previous dis-play:

2, 5, x1/2, y2, y3/2

Within each group of factors having the same base, we eliminate allbut the one with the highest exponent to obtain

2, 5, x1/2y2


Step 4: Multiply all the factors together left after step 3 toobtain the least common denominator. According to my LCD“algorithm,” the least common denominator is

LCD = (2)(5)x1/2y2 = 10x1/2y2.

Now combine!2√

x

5y2 +3

2x1/2y3/2 =2x1/2

5y2

[2x1/2

2x1/2

]+

32x1/2y3/2

[5y1/2

5y1/2

]

=4x

10x1/2y2 +15y1/2

10x1/2y2


2√

x

5y2 +3

2x1/2y3/2 =4x + 15y1/2

10x1/2y2

And that’s the procedure! Do you think you can do it?Example 4.3.

Important Points


If you erred on this one, more than likely it was on the appropriatemultiplicative constant: 6 not 18. At least that’s what I’m betting on.

The instructions of the LCD Algorithm said to completely factor thedenominator. Here’s a list of the factors

3, x3/2, y2,︸︷︷︸first term

2, 3, x, y4︸︷︷︸second term

Let’s rearrange them

2, 3, 3, x, x3/2, y2, y4

Now drop duplicate factors—that’s the 3. Oops! I did mention drop-ping identical factors, didn’t I?

2, 3, x, x3/2, y2, y4

Now, from each group all members of which have the same base, dropall but the one with the highest power.

2, 3, x3/2, y4


and the LCD is the product of same:

LCD = (2)(3)x3/2y4 = 6x3/2y4.

Alternative (a) will work as a common denominator, but it is not theleast common denominator. If you use (a), you will be working withlarger numbers than is really necessary. Important Point

��mptiimenu


Lesson 5: Expansion

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 5: Expansion

Table of Contents5. Expansion5.1. General Methods of Expansion

• Special Products • Radicals Revisited5.2. The Binomial Formula

• The Binomial Expansion Algorithm

5. ExpansionIn this lesson we discuss and illustrate methods of expanding a prod-uct of algebraic expressions. Initially, we discuss General Methodsof Expansion. Following that, we consider the problem of computingpowers of a binomial. Such expansion can be carried out quickly andquietly by using the Binomial Formula.

The student will find these techniques to be invaluable tools in theirstudy of algebra, calculus, and beyond . . . perhaps into an engineeringdiscipline.

5.1. General Methods of Expansion

In Lesson 4, in the section entitled The Distributive Law, welooked at products of the form a(b + c) and saw that

a(b + c) = ab + ac. (1)

When read from left-to-right, the formula is a rule for expanding aproduct. When read from right-to-left, the above formula can be usedfor simple factoring or combining of similar terms.

Section 5: Expansion

In this lesson we take up the problem of expanding more complicatedexpressions than we considered in Lesson 4. These more complicatedproducts are products in which both factors are the sum of severalterms: such as (x2 +2)(3x − 4), (a+ b)(a+ b+ c), or (x+1)2(x+2)3.

An algebraic expression consisting of exactly two terms is called abinomial. Consider the problem of computing the product of two bi-nomials: (a + b)(c + d). We can and do expand this product by theDistributive Law.

(a + b)(c + d ) = a · (c + d ) + b · (c + d ) � by (1)

= ac + ad + bc + bd � by (1)Thus,

General Multiplication Rule:

(a + b)(c + d) = ac + ad + bc + bd (2)


Key Point. If you study the formula (2), you can see that the productof two binomials is the sum of all possible products obtained by takingone term from the first factor and one term from the second factor.This observation is valid even when there are an arbitrary numberof factors in the product and an arbitrary number of terms in eachfactor.

The above observation then eliminates the need to memorize for-mula (2)! Let’s go the examples.

Example 5.1. Expand and combine each of the products.(a) (x + 1)(x + 2) (b) (2w − 3s)(5w + 2s)(c) (2x − 3)(x2 − 2) (d) (

√a +

√b)(

√a − √

b)

Study the above examples. Try to “see” the pattern and get a “feel”for multiplying out binomials.

Exercise 5.1. Expand and combine each of the following.(a) (4x − 5)(3x + 2) (b) (x − 2y)2 (c) (x3/2 + 1)(x1/2 + 2)


Now let’s look at multiplying binomials and trinomials (three terms).

(a + b)(x + y + z) = a(x + y + z) + b(x + y + z)= ax + ay + az + bx + by + bz

Thus,

(a + b)(x + y + z) = ax + ay + az + bx + by + bz

Notice that every term of the first factor is multiplied by every termof the second factor. This is key to understanding how to expandarbitrary complex expressions.

Example 5.2. Expand and combine each of the following.(a) (x + y)(x2 − xy + y2) (b) (2x2 − 4y3)(6x3 + 3xy + 2y2)

Practice the process of expanding on the next set of problems.

Exercise 5.2. Expand and combine each of the following.(a) (2x − y)(x2 − 2y2 + xy) (b) (ab − c)(a − b + c)


There is a perhaps a more convenient method for multiplying outsums. The following example illustrates the technique.

Example 5.3. Expand and combine (2x − 3y)(x2 − y2 + 2xy).

Exercise 5.3. Use the techniques in Example 5.3 to expand andcombine the expression (4x − 7y)(2x2 − 3y2 − 4xy).

• Special ProductsThere are many special products that I could present to you at thistime. Here are a few of the more important ones.

Some Special Products:� (x − y)(x + y) = x2 − y2 (3)

� (x + y)2 = x2 + 2xy + y2 (4)

� (x − y)2 = x2 − 2xy + y2 (5)

� (x + a)(x + b) = x2 + (a + b)x + ab (6)


Each of these equations is just a special case of the General Multipli-cation Rule; even so, these formulae speed up the expansion of thesecommon types of products and are, therefore, quite useful to know.

Each of the product rules has a verbalization: Just click on the greentriangle to jump there.

In the next series of examples we illustrate each of the above specialproduct rules. Use their verbalizations to help you understand howto use the formula—the verbalizations are independent of the lettersbeing used in the expansions and are very useful for that reason.

Illustration 1. Expand each of the following.(a) Equation (3) tells us that the product of the sum and differ-

ence of two expression is the difference of the squares of the twoexpressions. This formula has a verbalization which you shouldrecite every time you use the formula. This will help you remem-ber these simple expansion formulae.

1. (x2 − y3)(x2 + y3) = (x2)2 − (y3)2 = x4 − y6.2. (

√2 − 1)(

√2 + 1) = (

√2)2 − 1 = 2 − 1 = 1.


3.(x2

2+

1√3

)·(x2

2− 1√

3

)=

x4

4− 1

3.

4. (x − y)2(x + y)2 = [(x − y)(x + y)]2 = [x2 − y2]2.(b) Equations (4) and (5) tells us how to square a binomial. Each

have verbalizations and when you square you simply verbalizeas you expand. The verbalizations are: (4) and (5).

1. (2x + y)2 = 4x2 + 4xy + y2.2. (x − y)2(x + y)2 = [(x − y)(x + y)]2 = [x2 − y2]2 = x4 −

2x2y2 + y4.

3.(x

3− 3

x

)2=

x2

9− 2

(x

3

)( 3x

)+

9x2 =

x2

9− 2 +

9x2 .

(c) The previous formulae are just special cases of the general ex-pansion formula for binomials. Here are some examples of (6)—ittoo has a verbalization.

1. (x + 1)(x + 3) = x2 + (1 + 3)x + 3 = x2 + 4x + 3.2. (x + 3)(x − 4) = x2 + (3 − 4)x − 12 = x2 − x − 12.


3. (y − 1)(y + 2) = y2 + (−1 + 2)y − 2 = y2 + y − 2. Moregeneral products such as (4x − 3)(9x + 2) can be carriedout by the General Multiplication Rule.

Illustration Notes: Notice how in example (b), part 2, the expansionof the more complicated expression was easily accomplished by firstmultiplying the two bases together, as is permitted by Law #2, thenthe special product formula (3) was applied followed by (5). A lot ofideas were used in making this little calculation.Here is a couple of sets of problems for your consideration. Work themout first before you look at the solutions. If you missed some of them,study the solutions to understand what went wrong.

Exercise 5.4. Expand and combine each of the following.(a) (3x2 − 8y4)(3x2 + 8y4) (b) (xy +

√x)(xy − √

x)(c) (x−3 − 2x−2)(x−3 + 2x−2)

Exercise 5.5. Expand and combine each of the following.(a) (3x − 2)2 (b) (6y2 + 1)2 (c) (4 − √

2)2 (d) (2x2y3 − 3)2


Those were so easy (and important), let’s have more. If you erred,study the solutions and test you understanding again.

Exercise 5.6. Expand and combine each of the following.(a) (5 − 3x4)2 (b) (

√5 +

√6)2 (c) (x−3/2 − x−1/2)2

(d) (4x6y4 + 5)2

• Radicals RevisitedWe can exploit multiplication formula (3) in order to Rationalize theDenominator the denominators in certain situations.

If we are working with a ratios of the form

33 +

√2

or2√

5 − √2

(7)

(or some other variation on the same theme), we can rid ourselvesof the dastardly radicals in the denominator by multiplying by theconjugate of the denominator.


What is a conjugate? The conjugate of√5 − √

3 is√5 +

√3. The

conjugate of 3+√2 is 3−√

2. In general, if we have a sum or differenceof two terms, change the sign of the second term to get the conjugate.

Example 5.4. Rationalize the following two expressions.

(a)3

3 +√2

(b)2x√

5 − √3

(c)3 − √

23 +

√2

Try a few simple one’s yourself please. Study the level of simplificationofExample 5.4 and strive to attain the same level of simplicity. Whensolving all these problems, be neat and organized.

Exercise 5.7. Rationalize the denominator.

(a)7

4 − √2

(b)6√

3(√3 − 1)

(c)2 +

√5

2 − √5

5.2. The Binomial Formula

Let us turn now the problem of computing higher powers of binomials.In this section, we are interested in learning how to compute arbitrary


powers of a binomial:

Problem: Compute (a + b)n, for n= 1, 2, 3, 5, . . .

Let’s list out the first few cases:• (a + b)2 = a2 + 2ab + b2

• (a + b)3 = a3 + 3a2b + 3ab2 + b3

• (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.• (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4b5.• (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6.

The symbolic factors follow a definite pattern. In the expansion of(a + b)n the symbolic factors are

anb0, an−1b, an−2b2, an−3b3, . . . , a2bn−2, abn−1, a0bn

Pattern of the Symbolic Factors: Notice the power of a begin at nand decrease to power 0; the powers of b begins at 0 and increases topower n.


Pattern of the Numerical Coefficients: The pattern of the coefficientsis a little harder to see. There is a general formula for these coeffi-cients, called binomial coefficients, but that formula will not be pre-sented here. Instead, an algorithm for expanding a binomial will beemphasized.

• The Binomial Expansion AlgorithmHere is a series of steps that will enable you to expand without mem-orizing formulas.

Algorithm for expanding (a + b)n.(a) The first term is an. Call this the current term.(b) plus . . .

1. the product of the coefficient of the current term and thecurrent exponent of a, divided by the term number of thecurrent term, times . . .

2. a raised to one less power, times . . .3. b raised to one greater power.


(c) If the exponent of a just computed is zero then you are done.If the exponent of a is not zero, then call the term you justcomputed the current term and go to step (b).

The next example is the binomial algoritm in action. Study this ex-ample very closely! The same reasoning can be used to expand anybinomial.

Example 5.5. Illustrate this algorithm by using it to expand (a+b)3.

If you survived the reading of Example 5.5, and you have a basic“feel” for the algorithm, you know it is easier to expand out the termthan it is to type out a detailed explanation how to do it!

It is important to realize that the binomial algorithm is set up forexpanding basic symbols like a and b. When you want to expandsomething like (2x+3y)3, you expand using the algorithm with a = 2xand b = 3y. Thus,

(2x + 3y)3 = (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3. (8)


The powers of the first term decrease and the powers of the secondterm increase. The coefficients are computed by multiplying the cur-rent coefficient by the current exponent of the ‘a’ and dividing by theterm number.

Further simplification of (8) is necessary:

(2x + 3y)3 = 8x3 + 36x2y + 54xy2 + 27.

Example 5.6. Expand (2x2 − y3)4.

The binomials can be expanded by the algorithm, by memorizing thefirst so many expansions—as listed above, by using Pascal’s Trian-gle to compute the coefficients or by applying the Binomial coeffi-cient formula. The latter two are not presented here. :-( DPSThe expansion techniques is quite general and can be applied withfractional exponents or negative exponents.

Example 5.7. Expand (x1/2 − 1)3.


Let’s finish off this section with some exercises.

Exercise 5.8. Expand each of the following.(a) (4x − 3y)2 (b) (x1/2 − x−1/2)2 (c) (sin(x) − cos(x))2

Exercise 5.9. Expand each of the following.(a) (x − 2y)3 (b) (x3 + y5)4 (c) (x1/2 − 1)5

Tip. Perhaps you may have noticed that when expanding a difference(a − b)n the signs alternate. This observation accelerates the processof expansion.

Illustration 2. Expand (2x2 − x−3)3.

(2x2 − x−3)3 = (2x2)3 − 3(2x2)2(x−3) + 3(2x2)(x−3)2 − (x−3)3

= 8x6 − 12x4x−3 + 6x2x−6 − x−9

= 8x6 − 12x + 6x−4 − x−9

Additional simplifications are possible, but I’ll call it quits.

And now for the last exercise of this lesson!


Exercise 5.10. Expand each using alternating signs as a short-cut.Passing grade is 100%.(a) (3x4 − 2)3 (b) (x2y3 − 1)4

This is the end of Lesson 5. Click on Lesson 6 to continue.

Verbalizations

The Product of a sum and difference:

(x − y)(x + y) = x2 − y2

The product of the sum and difference of x and y is thesquare of the first minus the square of the second.

Verbalizations

Squaring a Binomial :

(x + y)2 = x2 + 2xy + y2

The square of a binomial is the square of the first, plus twicethe product of the first and second, plus the square of thesecond.

Verbalizations

Squaring a Binomial :

(x − y)2 = x2 − 2xy + y2

The square of a binomial is the square of the first, minustwice the product of the first and second, plus the square ofthe second.

Verbalizations

The Product of Two Binomials:

(x + a)(x + b) = x2 + (a + b)x + ab

The product of two binomials is the product of the two firstterms, plus the sum of the two cross-product terms, plus theproduct of the second terms.


5.1. Solutions:

(a) Expand and combine (4x − 5)(3x + 2).

(4x − 5)(3x + 2)

= (4x)(3x) + (4x)(2) + (−5)(3x) + (−5)(2)

= 12x2 + 8x − 15x − 10

= 12x2 − 7x − 10

(b) Expand and combine (x − 2y)2.

(x − 2y)2 = (x − 2y)(x − 2y)

= x2 + x(−2y) + (−2y)(x) + (−2y)(−2y)

= x2 − 2xy − 2xy + 4y2

= x2 − 4xy + 4y2


(c) Expand and combine (x3/2 + 1)(x1/2 + 2).

(x3/2 + 1)(x1/2 + 2)

= (x3/1)(x1/2) + (x3/2)(2) + (1)(x1/2) + 2

= x + 2x3/2 + x1/2 + 2

Optionally, the result can be left in radical notation:

(x3/2 + 1)(x1/2 + 2) = x + 2x√

x +√

x + 2

And, if we wanted to be true to our algebraic roots, we couldwrite

(x3/2 + 1)(x1/2 + 2) = x + (2x + 1)√

x + 2

Exercise 5.1.


5.2. Solutions:(a) Expand and combine (2x − y)(x2 − 2y2 + xy).

(2x − y)(x2 − 2y2 + xy)

= 2x(x2) − 2x(2y2) + 2x(xy) − y(x2) + y(2y2) − y(xy)

= 2x3 − 4xy2 + 2x2y − x2y + 2y3 − xy2

= 2x3 + (2x2y − x2y) + (−4xy2 − xy2) + 2y3

= 2x3 + x2y − 5xy2 + 2y3

(b) Expand and combine (ab − c)(a − b + c).

(ab − c)(a − b + c)

= ab(a) + ab(−b) + ab(c) − c(a) − c(−b) − c(c)

= a2b − ab2 + abc − ac + bc − c2

In this last problem, no additional simplification is necessary.Exercise 5.2.


5.3. Solution: Put the trinomial on top and the binomial on bottom.

2x2 − 3y2 − 4xy4x − 7y8x3 − 12xy2 − 16x2y

+ 28xy2 − 14x2y + 21y3

8x3 + 16xy2 − 30x2y + 21y3 Exercise 5.3.


5.4. Solutions:(a) (3x2 − 8y4)(3x2 + 8y4) = 9x4 − 64y8.(b) (xy +

√x)(xy − √

x) = x2y2 − x.(c) (x−3 − 2x−2)(x−3 + 2x−2) = x−6 − 4x−4 = x−6(1 − 4x2) =

1 − 4x2

x6 .

Study the last solution . . . several facts about factoring out and neg-ative exponents were used. Exercise 5.4.


5.5. Solutions: The square of a binomial is the square of the first(term) plus/minus twice the product of the first and second (terms),plus the square of the second (term). Apply that to each.

(a) (3x − 2)2 = 9x2 − 12x + 4 .

(b) (6y2 + 1)2 = 36y4 + 12y2 + 1 .

(c) (4 − √2)2 = 16 − 8

√2 + 2 = 18 − 8

√2 .

(d) (2x2y3 − 3)2 = 4x4y6 − 12x2y3 + 9 .

Short and sweet! DPS Exercise 5.5.


5.6. Solutions:

(a) (5 − 3x4)2 = 25 − 30x4 + 9x8 = 9x8 − 30x4 + 25.

(b) (√5 +

√6)2 = 5 + 2

√5

√6 + 6 = 11 + 2

√30.

(c) Expand (x−3/2 − x−1/2)2.(x−3/2 − x−1/2)2 = x−3 − 2x−3/2x−1/2 + x−1

= x−3 − 2x−2 + x−1

= x−3(1 − 2x + x2)

=x2 − 2x + 1

x3 � get rid of neg. exp.

• The other option to expanding this particular expression is therid yourself of the negative exponents and then square. Try solv-ing this problem this way as well. Check out the solution byclicking on the green dot.

(d) (4x6y4 + 5)2 = 16x12y8 + 40x6y4 + 25.Exercise 5.6.


5.7. Solutions:

(a) Rationalize7

4 − √2.

74 − √

2=

74 − √

2· 4 +

√2

4 +√2

� mul. by conjugate

=7(4 +

√2)

(4 − √2)(4 +

√2)

� sum times difference

=7(4 +

√2)

16 − 2� by (3)

=7(4 +

√2)

14

=12(4 +

√2)


(b) Rationalize6√

3(√3 − 1)

.

6√3(

√3 − 1)

=6

3 − √3

=6

3 − √3

· 3 +√3

3 +√3


=6(3 +

√3)

(3 − √3)(3 +

√3)

� sum times diff.

=6(3 +

√3)

(9 − 3)� by (3)

=6(3 +

√3)

6� cancel the 6’s!

= 3 +√3 � interesting!


(c) Rationalize2 +

√5

2 − √5.

2 +√5

2 − √5=

2 +√5

2 − √5

· 2 +√5

2 +√5


=(2 +

√5)(2 +

√5)

(2 − √5)(2 +

√5)


=(2 +

√5)2

4 − 5� by (3)

=4 + 4

√5 + 5

−1� from (5)

= −(9 + 4√5)

Exercise 5.7.


5.8. Solutions:

(a) (4x − 3y)2 = 16x2 + 2(4x)(−3y) + 9y2 = 16x2 − 24xy + 9y2.

(b) (x1/2 − x−1/2)2 = x − 2(x1/2)(x−1/2) + x−1 = x − 2 + x−1.

(c) Expand (sin(x) − cos(x))2.

(sin(x) − cos(x))2 = sin2(x) − 2 sin(x) cos(x) + cos2(x)

= (sin2(x) + cos2(x)) − 2 sin(x) cos(x)

= 1 − 2 sin(2x)

I tossed in this problem to shake you up a little. Here, I have recalledthe identities: sin2(x) + cos2(x) = 1 and 2 sin(x) cos(x) = sin(2x).

Exercise 5.8.


5.9. Solutions: Just expand each by the binomial algorithm. Notethat as I type out these solutions, I am expanding by the algorithm—I’m not using a list of formulas.(a) Expand (x − 2y)3

(x − 2y)3 = x3 + 3x2(−2y) + 3x(−2y)2 + (−2y)3

= x3 − 6x2y + 12xy2 − 8y3

(b) Expand (x3 + y5)4.

(x3 + y5)4 = (x3)4 + 4(x3)3(y5) + 6(x3)2(y5)2

+ 4(x3)(y5)3 + (y5)4

= x12 + 4x9y5 + 6x6y10 + 4x3y15 + y20


(c) Expand (x1/2 − 1)5

(x1/2 − 1)5 = (x1/2)5 + 5(x1/2)4(−1) + 10(x1/2)3(−1)2

+ 10(x1/2)2(−1)3 + 5(x1/2)(−1)4 + (−1)5

= x5/2 − 5x2 + 10x3/2 − 10x + 5x1/2 − 1

These terms can be converted to radical notation. Make it so!Exercise 5.9.


5.10. Solutions:(a) (3x4 − 2)3.

(3x4 − 2)3 = (3x4)3 − 3(3x4)2(2) + 3(3x4)(2)2 − 23

= 27x12 − 54x8 + 36x4 − 8

(b) (x2y3 − 1)4.

(x2y3 − 1)4 = (x2y3)4 − 4(x2y3)3 + 6(x2y3)2

− 4(x2y3) + 1

= x8y12 − 4x6y9 + 6x4y6 − 4x2y3 + 1

Exercise 5.10.


5.1. Solutions: In each case, we add up all possible products of termsfrom the first factor with terms in the second factor.(a) Expand and combine (x + 1)(x + 2).

(x + 1)(x + 2)

= (x)(x) + (x)(2) + (1)(x) + (1)(2) � Expand by (2)

= x2 + 2x + x + 2

= x2 + 3x + 2 � and combine!

(b) Expand and combine (2w − 3s)(5w + 2s).(2w − 3s)(5w + 2s)

= (2w)(5w) + (2w)(2s) + (−3s)(5w) + (−3s)(2s)

= 10w2 + 4sw − 15sw − 6s2

= 10w2 − 11sw − 6s2


(c) Expand and combine (2x − 3)(x2 − 2).(2x − 3)(x2 − 2)

= (2x)(x2) + (2x)(−2) + (−3)(x2) + (−3)(−2)

= 2x3 − 4x − 3x2 + 6

= 2x3 − 3x2 − 4x + 6

Notice how the negative terms are handled during the expansionof this product. This would be a good style for you.

(d) Expand and combine (√

a +√

b)(√

a − √b).

(√

a +√

b)(√

a −√

b)

=√

a (√

a ) +√

a (−√

b ) +√

b (√

a ) +√

b (−√

b )

= a − √a√

b +√

b√

a − b

= a − b

In this calculation, the so-called cross-product terms eliminateeach other.

Example 5.1.


5.2. Solutions: We multiply each term of the first factor by each termof the second factor, and add together the resultant calculations.

(a) Expand and combine (x + y)(x2 − xy + y2).

(x + y)(x2 − xy + y2)

= x(x2) + x(−xy) + x(y2) + y(x2) + y(−xy) + y(y2)

= x3 − x2y + xy2 + x2y − xy2 + y3

= x3 + y3

(b) Expand and combine (2x2 − 4y3)(6x3 + 3xy + 2y2).

(2x2 − 4y3)(6x3 + 3xy + 2y2)

= 2x2(6x3) + 2x2(3xy) + 2x2(2y2)

− 4y3(6x3) − 4y3(3xy) − 4y3(2y2)

= 12x5 + 6x3y + 4x2y2 − 24x3y3 − 12xy4 − 8y5

Example 5.2.


5.3. This technique is similar to multiplying out numbers by penciland paper.

x2 − y2 + 2xy2x − 3y2x3 − 2xy2 + 4x2y

− 6xy2 − 3x2y + 3y3

2x3 − 8xy2 + x2y + 3y3

The third row is obtained by taking the first term in the second rowand multiplying it by each term in the first row.

The fourth row is obtained by taking the second term in the secondrow and multiplying it by each term in the first row, being sure toplace similar terms in the same column. Example 5.3.


5.4. Solutions:

(a)3

3 +√2.

33 +

√2=

33 +

√2

· 3 − √2

3 − √2


=3(3 − √

2)(3 +

√2)(3 − √

2)� sum times difference

=3(3 − √

2)9 − 2

� by (3)

=3(3 − √

2)7

=37(3 −

√2)


(b)2x√

5 − √3.

2x√5 − √

3=

2x√5 − √

3·√5 +

√3√

5 +√3


=2x(

√5 +

√3)

(√5 − √

3)(√5 +

√3)


=2x(

√5 +

√3)

5 − 3� by (3)

=2x(

√5 +

√3)

2

= x(√5 +

√3 )


(c)3 − √

23 +

√2.

3 − √2

3 +√2=

3 − √2

3 +√2

· 3 − √2

3 − √2


=(3 − √

2)(3 − √2)

(3 +√2)(3 − √

2)� sum times difference

=(3 − √

2)2

9 − 2� by (3)

=9 − 6

√2 + 2

7� from (5)

=11 − 6

√2

7

Example 5.4.


5.5. Expand (a + b)n using the binomial algorithm.1. The first term is a3. This is the current term.� (a + b)3 = a3 + . . .2. plus . . .

a. the product of the coefficient of the current term (1) andthe current exponent of a (3), divided by the current termnumber (1): (1)(3)/1 = 3 times . . .

b. a raised to one less power: 3a2 times . . .c. b raised to one greater power: 3a2b

3. Thus, the second term is 3a2b. This is our new current term.The expansion so far is . . .

� (a + b)3 = a3 + 3a2b + . . .4. The exponent of a is nonzero so we repeat step (b).5. plus . . .

a. the product of the coefficient of the current term (3) andthe current exponent of a (2), divided by the current termnumber (2): (3)(2)/(2) = 3 times . . .

b. a raised to one less power: 3a times . . .c. b raised to one greater power: 3ab2.


6. Thus, the third term is 3ab2. This is our new current term. Theexpansion so far is . . .

� (a + b)3 = a3 + 3a2b + 3ab2 + . . .7. The exponent of a is nonzero so we repeat step (b).8. plus . . .

a. the product of the coefficient of the current term (3) andthe current exponent of a (1), divided by the current termnumber (3): (3)(1)/(3) = 1 times . . .

b. a raised to one less power: (1)a0 times . . .c. b raised to one greater power: (1)a0b3 = b3.

9. Thus, the fourth term is b3. This is our current term.� (a + b)3 = a3 + 3a2b + 3ab2 + b3

10. The exponent of a is down to zero! Finished!

We have expanded (a + b)3 as

(a + b)3 = a3 + 3a2b + 3ab2 + b3.

And that’s all there is to it! Example 5.5.


5.6. We first use the binomial algorithm to obtain the form of theexpansion. Note that a = 2x2 and b = −y3.

(2x2 − y3)4 = (2x2 + (−y3))4

= (2x2)4 + 4(2x2)3(−y3) + 6(2x2)2(−y3)2

+ 4(2x2)(−y3)3 + (−y3)4

Now we simplify from there:

(2x2 − y3)4 = 16x8 − 32x6y3 + 24x4y6 − 8x2y9 + y12

The student, that’s you, should verify the above simplification.Example 5.6.


5.7. Expand the cubic using the algorithm with a = x1/2 and b = 1:

(x1/2 − 1)3 = (x1/2)3 + 3(x1/2)2(−1) + 3(x1/2)(−1)2 + (−1)3

= x3/2 − 3x + 3x1/2 − 1


(x1/2 − 1)3 = x3/2 − 3x + 3x1/2 − 1.

Or we can take a radical approach to presentation the answer.

(√

x − 1)3 = x√

x − 3x + 3√

x − 1.

Example 5.7.

Important Points


Alternate Solution: Expand (x−3/2 − x−1/2)2.

(x−3/2 − x−1/2)2 =(x−3/2(1 − x)

)2� factor out lowest power

= (x−3/2)2(1 − x)2 � Law #2

= x−3(1 − 2x2 + x2) � Law #3 & (5)

=x2 − 2x + 1

x3

Same expansion as before! No surprise. Important Point

��

mptiimenu


Lesson 6: Dividing & Factoring Polynomials

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 6: Dividing & Factoring Polynomials

Table of Contents6. Dividing & Factoring Polynomials6.1. Polynomials: A Quick Review6.2. Polynomial Division

• Polynomial Division Explained6.3. Factoring Polynomials: Motivation6.4. Factoring Polynomials: Theory

• Reducible versus Irreducible Polynomials • The Funda-mental Theorem of Algebra • Roots and Linear Factors Re-lated

6.5. Factoring Polynomials: Methods• Factoring x2 + bx + c • Factoring x2 − a2 • Factoringax2 + bx + c • Factoring x3 ± a3

6. Dividing & Factoring Polynomials

6.1. Polynomials: A Quick Review

A polynomial in x is an algebraic expression that can be built upthrough any (finite) combination of additions, subtractions, and mul-tiplications of the symbol x with itself and with numerical constants.The degree of the polynomial is the value of the highest exponent.

Illustration 1. Examples of Polynomials in x.(a) 2x+1 has degree 1. A degree one polynomial is sometimes called

linear because its graph is a straight line.(b) 5x2 − 4x + 3 has degree 2. A degree two polynomial is called a

quadratic polynomial.

(c) 7x3 − 4x2 − 23x +

89

has degree 3. This is called a cubic poly-nomial.

(d) x4 + x has degree 4.(e) x45 − 5x34 + x3 + 1 has degree 45.(f) −3y3 + 2y − y + 1

2 is a polynomial in y of degree 3.

Section 6: Dividing & Factoring Polynomials

(g) As a convenience, a constant is considered to be a polynomialof degree 0. Thus, the algebraic expression 3 may be interpretedas a polynomial of degree 0.

Sometimes we symbolically denote a polynomial in x by notationssuch as P (x) or Q(x); polynomials in some other variable such as ywould be denoted similarly: P (y) and Q(y). If P (x) is a polynomialin x, then we write (type) deg(P (x)) to refer to the degree of P (x).

Polynomials arise naturally in many branches of mathematics andengineering. Polynomials are the basic building blocks used to createnumerical approximations such as the ones used by your hand-heldcalculator. The graphs of certain polynomials have properties thatman exploits to create many useful everyday conveniences such asflashlights and satellite dishes.

One of the reasons for the importance of polynomials—and quotientsof polynomials—is that their values can be computed by elementary


arithmetic operations: addition, subtraction, multiplication and di-vision. These are the operations a computer is designed to performquickly and efficiently.

The sum and product of polynomials is again a polynomial. Illustratethis assertion by doing the next exercise. Classify each “answer” as apolynomial and state its degree.

Exercise 6.1. The Algebra of Polynomials. Perform the indi-cated operations and classify the results.(a) (4x3 − 6x2 + 2x + 1) + (2x3 − 3x + 4)(b) (7x5 − 4x3 + 12x − 4) − (5x5 + 3x4 + 4x3 + 2x − 4)(c) (3x − 2)(x2 − 2x + 1)(d) (x2 − 4)(x2 + 4)

Adding and multiplying polynomials were covered by the generalmethods described in Lesson 3 (addition) and in Lesson 5 (mul-tiplication). In the next paragraph we look at division of polynomials.


6.2. Polynomial Division

In this paragraph, we will be primarily interested in dividing polyno-mials. If N(x) and D(x) are polynomials in x then the expression

N(x)D(x)

(1)

is called a rational expression or a rational function.

Illustration 2. Examples of Rational Expressions.

(a)2x + 1

5x2 − 4x + 2(b)

3x3 − 2x2 + 7x + 1x2 + 3x − 2

(c)4x12 − 7x9 + 1

x3 (d)4x4 − 3x3 + 2x2 + 8x − 3

x + 3


• Polynomial Division ExplainedNow turning to the problem of computing (1), generally, we can onlydivide a ratio of polynomials when the when the degree of the nu-merator is greater than or equal to the degree of the denominaotr. Insymbols

If deg(N) ≥ deg(D), then we can divideN(x)D(x)

In this case, when we do divide, then the result looks like this

N(x)D(x)

= Q(x) +R(x)D(x)

(2)

where Q(x) and R(x) are polynomials and deg (R(x)) < deg(D(x)).An interesting and important fact to remember is that the represen-tation on the right-hand side of equation (2) is unique. This fact willbe exploited in the section on factoring.

Terminology. Given the representation in equation (2), then• N(x) is called the dividend;• D(x) is called the divisor;


• Q(x) is called the quotient;• R(x) is called the remainder.

This is the same terminology used in long division of numbers.

In Illustration 2, the numerator has less degree than the denomi-nator in example (a); consequently, we cannot divide these two poly-nomials. In examples (b)–(d), the degree of the numerator is equal toor greater than the degree of the denominator; in each of these exam-ples we can further expand the expression by dividing denominatorinto the numerator.

Division Algorithm. How to divide one polynomial by another andobtain a result of the form (2)(a) Arrange the terms of N(x) and D(x) so that the powers of the

terms are listed in descending order.(b) Divide the first term of the dividend by the first term of the

divisor. This gives the first term of the quotient.


(c) Now, multiply the term of the quotient just computed by thedivisor, and subtract this product from the dividend. The resultis the remainder.

� If the degree of the remainder is less than the degree ofthe divisor, D(x), you are done!

� If the degree of the remainder is not less than the divisor,D(x), continue by using the remainder just obtained as anew dividend, repeat; i.e., go to (b) to compute the nextterm of the quotient.

The next example shows how to perform polynomial division. Studythis example closely; with a paper and pencil, slowly work throughthe example—follow my calculations as I work through the DivisionAlgorithm.

Example 6.1. Divide the expression:2x3 − 3x2 + 6x − 4

x − 1.


Again, work through these two examples with pencil and paper. Striveto understand how each entry is determined from the Division Al-gorithm.

Example 6.2. Divide denominator into the numerator in parts (b)and (d) in Illustration 2.

Part (c) of Illustration 2 can be handled differently than parts (b)and (d). Why? Read the example to find out why.

Example 6.3. Divide4x12 − 7x9 + 1

x3 that appeared in part (c) ofIllustration 2.

Having done a few examples of polynomials, its time for you to try afew. Follow the Division Algorithm and my examples.

Exercise 6.2. Divide:2x3 − 3x2 + x − 1

x − 2.

Exercise 6.3. Divide:4x4 + 2x2 + x + 1

x2 + 1.


In the next exercise the remainder is zero. Do you recall what thatmeans?

Exercise 6.4. Divide:x3 + 1x + 1

.

6.3. Factoring Polynomials: Motivation

Factoring is the reverse process to expanding:

(x + 1)(x + 2) = x2 + 3x + 2. (3)

Reading the equation from left-to-right, we are expanding the poly-nomial (x + 1)(x + 2); reading the equation from right-to-left we arefactoring the polynomial x2 + 3x + 2.

Illustration 3. Factoring has a variety of uses in mathematics. Hereis a set of examples to illustrate that assertion.


(a) Simplification of algebraic expressions:

x3 + 3x2 + 2xx + 1

=x(x2 + 3x + 2)

x + 1� factor a common x

=x(x + 1)(x + 2)

x + 1� factor again by (3)

= x(x + 2) � nicely simplified!

Surely you agree, the last expression is preferable to the first.(b) Solving Equatons: For what values of x is x2+3x+2 equal zero?

That is, solve the equation

x2 + 3x + 2 = 0

Since x2 + 3x + 2 = (x + 1)(x + 2) it is now obvious that theonly values of x that make x2 + 3x + 2 equal zero are x = −1and x = −2.


(c) Simplification for the purpose of numerical calculations. Sup-pose you wanted to compute, on your calculator, values of thepolynomial

x4 + 4x3 + 6x2 + 4x+1

for x = 3.23344, −27.3234, and 34.000123. Even with the aid ofyour calculator this would be a tedious task, and there is a verygood chance that you will make errors. Suppose I told you thatthe above expression is nothing more than

(x + 1)4

Which one would you use to make the calculations?

Summary. Factorization tends to simplify, reduce the number of arith-metical operations performed, and often times yields valuable infor-mation about the behavior of the expression.


6.4. Factoring Polynomials: Theory

In this section we discuss some important background information andtheory: Reducible versus Irreducible Polynomials; the FundamentalTheorem of Algebra; and Roots and Factors Related.

These are important and fundamental concepts you should try tounderstand.

• Reducible versus Irreducible PolynomialsSome polynomials can be factored, others cannot. Naturally we havea terminology for each situation.

A polynomial that can be factored into a product of polynomials ofsmaller degree is called a reducible; otherwise, the polynomial iscalled irreducible.

Illustration 4. Reducible and Irreducible Polynomials.(a) The polynomial x2+3x+2 is reducible because it can be factored

into factors of smaller degree:

x2 + 3x + 2 = (x + 1)(x + 2).


We have factored a second degree polynomial into a product oflinear polynomials.

(b) The polynomial x3 − 3x2 + 3x − 1 is reducible because

x3 − 3x2 + 3x − 1 = (x − 1)3.

Here we say that (x − 1) is a linear factor of multiplicity 3.(c) The polynomial x3 + x2 + x + 1 is reducible because

x3 + x2 + x + 1 = (x + 1)(x2 + 1).

This polynomial factors into linear and quadratic factors.(d) The polynomials x2 + 1 and x2 − x + 1 are irreducible; that is,

it cannot be factored any further. How do I know that? Readon!

Question. It was stated in the last illustration that x2+1 is irreducible.Suppose I factor it as follows: x2 + 1 = 1

2 (2x2 + 2). Since I have

factored it into a product of polynomials (the constant 12 may be

considered a polynomial of degree 0), does this mean that x2 + 1 is


reducible? Review the definitions of reducible and irreducible beforeyou respond.(a) Yes (b) No

Question. Consider the factorization: 2x2 + 2x = (2)(x2 + x). Wehave factored a degree two polynomial into a product of a degree 0polynomial and a degree 2 polynomial. In light of the discussion inthe previous Question, does this mean that 2x2 + 2x is irreducible?(a) Yes (b) No

• The Fundamental Theorem of AlgebraAccording to the Fundamental Theorem of Algebra, any poly-nomial of degree greater than zero can be factored into a product oflinear and irreducible quadratic factors.

Illustration 5. For example,

x4 + 2x3 + 2x2 + 2x + 1 = (x + 1)2(x2 + 1). (4)


Linear Factors: The factor (x + 1)2 is called a linear factor becausex + 1, the base of the power, is a degree 1 polynomial (degree 1 =linear). The presence of (x+1)2 in the factorization, of course, meansthat (x + 1) appears twice in the factorization

x4 + 2x3 + 2x2 + 2x + 1 = (x + 1)(x + 1)(x2 + 1).

but hardly ever write it this way. Here, we say that (x+1) is a linearfactor of multiplicity 2, meaning it appears twice in the factorization.

Irreducible Quatdratic Factors: The other factor, (x2 + 1), in equa-tion (4) is a second degree polynomial, or a quadratic polynomial. Thisparticular one is irreducible. The exponent of the factor (x2 + 1) =(x2 + 1)1 is 1 meaning this factor has multiplicity 1; there is only 1factor of this type present in the factorization (4).

Thus, the factorization of the polynomial x4+2x3+2x2+2x+1 givenin equation (4) is the one described in the Fundamental Theoremof Algebra; that is,

x4 + 2x3 + 2x2 + 2x + 1 = (x + 1)2(x2 + 1).


is a factorization into linear and irreducible quadratics factors.

Exercise 6.5. Consider the factorization:

x3(2x + 1)4(x2 + 1)5(5 − 2x)3(x2 + x + 1).

Classify each factor as linear or irreducible quadratic and state themultiplicity of each factor.

• Roots and Linear Factors RelatedThere is a relationship between a root of a polynomial and its linearfactors. Let’s begin by recalling what a root of a polynomial is. Let

P (x) = ax3 + bx2 + cx + d

be any polynomial. (For illustrative purposes, I’ve just written a thirddegree polynomial.) A root of P (x) is any number r such that

P (r) = ar3 + br2 + cr + d = 0 (5)


that is, a root is any number r that causes the polynomial to evaluateto zero. Another way of thinking of a root is, a root is any solution tothe equation:

ax3 + bx2 + cx + d = 0.

Illustration 6. The numbers −1 and −2 are roots of the polynomialP (x) = x2 + 3x + 2 since

� P (−1) = (−1)2 + 3(−1) + 2 = 1 − 3 + 2 = 0

� P (−2) = (−2)2 + 3(−2) + 2 = 4 − 6 + 2 = 0

In the next lesson, Lesson 7, we discuss techniques for finding rootsof polynomials. There, we will discuss methods of solving equations.

What is the relation between roots and linear factors?

� (x− r) a factor implies r is a root. Let P (x) be any polynomial andr a number. Clearly, if (x − r) is a factor or P (x), then r is a root ofP (x). Indeed, if we assume (x − r) is a factor of P (x) then

P (x) = Q(x)(x − r)


for some polynomial Q(x); that is, (x−r) appears in the factorizationof P (x). Now you can see that

P (r) = Q(r)(r − r) = 0.

This means that r is a root of the polynomial P (x). �

� r a root implies (x−r) a factor. Divide the polynomial P (x) by thepolynomial (x − r). The result, by equation (2) will have the form:

P (x)(x − r)

= Q(x) +R(x)

(x − r)(6)

where the remainder R(x) has degree less than x − r. This means, inthis case, that R(x) has degree zero since x−r had degree one. Degreezero polynomials are constants; thus, R(x) is, in fact, a constant. Illcall this constant R. We have then

P (x)(x − r)

= Q(x) +R

(x − r).


Multiply both sides of the equation by x − r to obtain

P (x) = Q(x)(x − r) + R, R = constant. (7)

There is a simple interpretation for the value of R. If we replace x byr in equation (7), we see

P (r) = Q(r)(r − r) + Ror,

R = P (r) (9)

Now, if r is a root of the polynomial P (x), then from (5), P (r) = 0.But by (9), the remainder R = P (r) = 0. Substituting this intoequation (7) we get,

P (x) = Q(x)(x − r) + 0, or, P (x) = Q(x)(x − r)

This means that (x − r) is a factor of P (x). �

We now summarize the observations of the past few paragraphs.


Summary. Roots and Linear Factors Related:Let P (x) is a polynomial and r a number. Then r is a rootof P (x) if and only if (x − r) is a factor of P (x).

I still haven’t explained how I know that x2 + 1 is irreducible. It isclear that for any number x, x2 + 1 �= 0; therefore, the polynomialx2 +1 has no roots, hence, it has no linear factors. Now x2 +1 hasdegree two, if it is to be factored, it must factor into linear factors.But x2 + 1 has no linear factors. Thus, x2 + 1 is irreducible.

Exercise 6.6. Read carefully the reasoning in the previous para-graph, and apply it to the polynomial x4 +1. It is clear that for all x,x4+1 �= 0 and so x4+1 has no roots, hence, has no linear factors. Canwe deduce that x4 + 1 is irreducible in the same way as we deducedthat x2 + 1 was irreducible?

6.5. Factoring Polynomials: Methods


• Factoring x2 + bx + cIn this section we discuss methods of factoring a second degree poly-nomial with integer coefficients with leading coefficient of one:

x2 + bx + c b, c ∈ Z

If the polynomial x2+bx+c is not irreducible, then it can be factoredinto a product of linear polynomials of the form:

x2 + bx + c = (x + r1)(x + r2). (10)

The problem is to determine the values of r1 and r2. This can bedone in one of two ways: (1) by trial and error; or (2) the QuadraticFormula. The latter method will be taken up in Lesson 7.

Let’s play for a moment. Expand the right-hand side of (10):

x2 + bx + c = (x + r1)(x + r2)

= x2 + (r1 + r2)x + r1r2.


Thus, if x2 + bx + c can be factored into (x + r1)(x + r2), then

r1 + r2 = b

r1r2 = c

Let’s now summarize the results so far in the form of a shadow box.

How to factor x2 + bx + c, where b, c ∈ Z

We try to find two number r1 and r2 such that

r1 + r2 = b

r1r2 = c

In this case, x2 + bx + c = (x + r1)(x + r2).

Strategy. Given x2 + bx + c, where b and c are integers: First list allpairs of integers r1 and r2 the product of which is c. Then among allthose pairs of numbers, r1 and r2, thus listed, choose the pair whosesum is b—choose that pair r1 and r2 such that r1 + r2 = b.


Let’s go to the examples.

Example 6.4. Factor each of the following polynomials.(a) x2 − x − 2 (b) x2 + 5x + 6 (c) 4 − 3x − x2

Solve the next exercise using good notation. Be neat, be organized.Take pride in your work !

Exercise 6.7. Factor each of the following using the recommendedstrategy as illustrated in Example 6.4.(a) x2 + 7x + 10 (b) x2 − 7x + 10 (c) x2 − 3x − 10(d) x2 + 3x − 10 (e) x2 + 11x + 10 (f) x2 − 9x − 10

Here’s a few more. Don’t forget the strategy and the standard methodsas illustrated in Example 6.4.

Exercise 6.8. Factor each of the following.(a) x2 + 4x − 12 (b) x2 + 3x − 18 (c) x2 − 10x + 21(d) x2 + 7x − 8 (e) x2 − 2x + 1 (f) 2x2 + 8x + 8


• Factoring x2 − a2

Polynomials that are in the form of a difference of two squares areeasy to factor. From equation (3) of Lesson 5 we have

x2 − a2 = (x − a)(x + a) (11)

The problem students have is recognizing the presence of such a poly-nomial. It’s a matter of training you eyes and brain to work together.

Illustration 7. Difference of Two Squares.(a) x2 − 1 = (x − 1)(x + 1).(b) x2 − 4 = (x − 2)(x + 2).(c) x2 − 3 = (x − √

3)(x +√

3). Note: Any positive number can bethought of as the square of another number. Consequently, thea2 that is in equation (11) does not have to be a ‘perfect square’;it can be any positive number.

Exercise 6.9. (Skill Level 0) Factor each of the following.(a) x2 − 9 (b) x2 − 12 (c) x2 − 17 (d) 25 − x2


Factorization formula (11) came from expansion formula (3) of Les-son 5. The expansion formula is actually more general: We can factorany difference of two squares—as illustrated below.

Illustration 8. Difference of two squares.(a) 4x2 − 9 = (2x − 3)(2x + 3).(b) 3x2 − 16 = (

√3x − 4)(

√3x + 4).

(c) x4 −16 = (x2 −4)(x2 +4) = (x−2)(x+2)(x2 +4). Here we haveapplied equation (11) twice! The result is a complete factoriza-tion of x4 −16 into a product of linear and irreducible quadraticfactors as prescribed by the Fundamental Theorem of Al-gebra.

Exercise 6.10. Factor each of the following using (11).(a) 4x2 − 9 (b) 5x2 − 3 (c) x4 − 25

The factorization formula (11) can be applied whenever we have adifference of two squares.

Exercise 6.11. Factor each of the following differences of squaresand simplify when possible. Note: ‘simplify’ �= ‘expand’.


(a) x2 − (x − 1)2 (b) x2y2 − 4 (c)(2x − 1)2 − (2x + 1)2

x

• Factoring ax2 + bx + cWe now turn to the problem of factoring a quadratic polynomial ofthe form

ax2 + bx + c a, b, c ∈ Z (12)

The trial and error methods illustrated earlier can be utilized to fac-tor (12). Generally, if the polynomial is not irreducible, its factoriza-tion of would look like this:

ax2 + bx + c = (q1x + r1)(q2x + r2) (13)

If we multiply out the right-hand side of (13) that will give us clueshow to find the numbers q1, q2, r1, and r2:

ax2 + bx + c = (q1x + r1)(q2x + r2)

= q1q2x2 + (q1r2 + r1q2)x + r1r2


Thus, we seek numbers such that

q1q2 = a r1r2 = c q1r2 + q2r1 = b

The task of finding the numbers q1, q2, r1, and r2 may be easy ordifficult.

Important Point. It is known that if the polynomial ax2 + bx + c hasonly rational roots—roots that are rational numbers—then q1, q2, r1,and r2 are integers. However, a quadratic polynomial such as (12)may have irrational roots.

In this section, we look at quadratics having rational roots, hencethe factors we seek have integer coefficients. The other case can behandled by use of the binomial formula taken up in Lesson 7.

Strategy. To factor ax2 + bx + c into to linear factors

ax2 + bx + c = (q1x + r1)(q2 + r2)

(a) List all integer pairs q1 and q2 such that q1q2 = a, the coefficientof the x2 term.


(b) List all integer pairs r1 and r2 such that r1r2 = c, the constantterm.

(c) Find that combination of q1, q2 and r1, r2 such that q1r2+q2r1 =b, the coefficient of the cross-product term.

This can usually be carried out by first choosing candidates for q1 andq2 then trying all combinations for r1 and r2. If that fails, try anotherchoice for q1 and q2. (I told you it was ‘trial and error.’)

Here’s an example of the scheme just described.

Example 6.5. Factor 6x2 + x − 1.

In the exercises below, use Example 6.5 as a guide to factoring.

Exercise 6.12. Factor 8x2 + 2x − 1.

Exercise 6.13. Factor 6x2 − 5x − 6.

In Lesson 7 we’ll review techniques of solving equations. Solvingpolynomial equations has applications to factorization. (Recall theparagraph on Roots and Linear Factors Related.) At that time we will


be able to factor quadratics with irrational roots as well as rationalroots.

• Factoring x3 ± a3

The sum or difference of cubes is easy to factor so we’ll finish thislesson by factoring this type. The formula we shall use is

b3 − a3 = (b − a)(b2 + ab + a2)

b3 + a3 = (b + a)(b2 − ab + a2)(14)

The major problem of students is recognizing the presence of cubes.

Illustration 9. Factor each of the following.(a) x3 − 8 = (x − 2)(x2 + 2x + 4). Here a = 2 in (14).(b) x3 + 8 = (x + 2)(x2 − 2x + 4).(c) 8x3 − 1 = (2x − 1)(4x2 + 2x + 1). Here b = 2x and a = 1.(d) x3y3 +1 = (xy+1)(x2y2 −xy+1). Here b = xy and a = 1.

Exercise 6.14. (Skill Level 0) Factor each of the following.(a) x3 − 1 (b) 27x3 − 8 (c) 8y6 + 27


Exercise 6.15. (Skill Level 0.5) Factor and simplify each.

(a)x3 − 1x − 1

(b)(8x3 − 1)(x2 + 3x + 2)

x2 − 1(c)

8x3 − 274x2 − 9

This is the end of Lesson 6 on dividing polynomials and factoring.To continue, go to Lesson 7.


6.1. Solutions:(a) Combine (4x3 − 6x2 + 2x + 1) + (2x3 − 3x + 4).

(4x3 − 6x2 + 2x + 1) + (2x3 − 3x + 4) = 6x3 − 6x2 − x + 5.

This is a polynomial of degree 3.(b) (7x5 − 4x3 + 12x − 4) − (5x5 + 3x4 + 4x3 + 2x − 4)

(7x5 − 4x3 + 12x − 4) − (5x5 + 3x4 + 4x3 + 2x − 4)

= 2x5 − 3x4 − 8x3 + 10x

which is a polynomial of degree 5.


(c) (3x − 2)(x2 − 2x + 1). Use the General Multiplication Rule toobtain

(3x − 2)(x2 − 2x + 1)

= (3x)(x2) − (3x)(2x) + (3x) − 2(x2) + 2(2x) − 2

= 3x3 − 6x2 + 3x − 2x2 + 4x − 2

= 3x3 − 8x2 + 7x − 2

Alternately, you can expand as follows:x2 − 2x + 1

3x − 23x3 − 6x2 + 3x

− 2x2 + 4x − 23x3 − 8x2 + 7x − 2 ⇐= The is a polynomial of degree 3.

(d) (x2 − 4)(x2 +4) = x4 − 16. by (3) in Lesson 5. A polynomialof degree 4.

Exercise 6.1.


6.2. Solution: Just follow the Division Algorithm!

2x2+ x + 3 � the quotient

x − 2 2x3−3x2+ x−1 � (b) divide 2x3 by x to get 2x2

2x3−4x2 � (c) multiply 2x2 by x − 2 and place it here

x2+ x−1 � (c) subtract; (b) divide x2 by x to get x

x2−2x � (c) multiply x by x − 2 and place it here

3x−1 � (c) subtract; (b) divide 3x by x to get 3

3x−6 � (c) multiply 3 by x − 2 and place it here

5 � (c) subtract; done; this is the remainder

Interpretation of Calculations: x − 2 divides into 2x3 − 3x2 + x − 1. . . 2x2 + x + 3 times with a remainder of 5. Thus,

2x3 − 3x2 + x − 1x − 2

= 2x2 + x + 3 +5

x − 2

Compare this equation with (2). Exercise 6.2.


6.3. Solution: Just follow the Division Algorithm!

4x2− 2 � the quotient

x2 + 1 4x4+2x2+x+1 � (b) divide 4x4 by x2 to get 4x2

4x4+4x2 � (c) multiply 4x2 by x2 + 1 and place it here

−2x2+x+1 � (c) subtract; (b) divide −2x2 by x2 to get −2

−2x2 −2 � (c) multiply −2 by x2 + 1 and place it here

x+3 � (c) subtract; done; this is the remainder

Interpretation of Calculations: x2 + 1 divides into 4x4 + 2x2 + x + 1. . . 4x2 − 2 times with a remainder of x + 3. Thus,

4x4 + 2x2 + x + 1x2 + 1

= 4x2 − 2 +x + 3x2 + 1

Compare this equation with (2). Exercise 6.3.


6.4. Solution:

x2−x +1 � the quotient

x + 1 x3 +1 � (b) divide x3 by x to get x2

x3+x2 � (c) multiply x2 by x + 1 and place it here

−x2 +1 � (c) subtract; (b) divide −x2 by x to get −x

−x2−x � (c) multiply −x by x + 1 and place it here

x+1 � (c) subtract; (b) divide x by x to get 1

x+1 � (c) multiply 1 by x + 1 and place it here

0 � (c) subtract; done; this is the remainder

Interpretation of Calculations: x+1 divides into x3 +1 . . . x2 −x+1times with a remainder of 0. Thus,

x3 + 1x + 1

= x2 − x + 1 +0

x + 1= x2 − x + 1


This means that x +1 evenly divides into x3 +1. If we multiply bothsides of the equations by x + 1 we get

x3 + 1 = (x + 1)(x2 − x + 1)

The division process can be used as a technique to factor a polynomial.Exercise 6.4.


6.5. Answers:Factor Classification Multiplicityx3 linear 3(2x + 1)4 linear 4(x2 + 1)5 irreducible quadratic 5(5 − 2x)3 linear 3(x2 + x + 1) irreducible quadratic 1

Exercise 6.5.


6.6. No! x4 + 1 has no linear factors, but, by the FundamentalTheorem of Algebra it can be factored into a product of linearfactors and irreducible quadratic factors. As it doesn’t have any linearfactors, and its degree is 4, it is must be possible to factor it into aproduct of two irreducible quadratic factors.

Can you find these two factors? Exercise 6.6.


6.7. Answers: Hopefully, you used the method illustrated in Exam-ple 6.4.(a) x2 + 7x + 10 = (x + 2)(x + 5).(b) x2 − 7x + 10 = (x − 2)(x − 5).(c) x2 − 3x − 10 = (x + 2)(x − 5).(d) x2 + 3x − 10 = (x − 2)(x + 5).(e) x2 + 11x + 10 = (x + 1)(x + 10).(f) x2 − 9x − 10 = (x + 1)(x − 10).

Did I list all possible combinations, or did I miss one or two?Exercise 6.7.


6.8. Answers:(a) x2 + 4x − 12 = (x − 2)(x + 6).(b) x2 + 3x − 18 =) = (x + 6)(x − 3).(c) x2 − 10x + 21 = (x − 3)(x − 7).(d) x2 + 7x − 8 = (x − 1)(x + 8).(e) x2 − 2x + 1 = (x − 1)(x − 1) = (x − 1)2—a perfect square!(f) 2x2+8x+8 = 2(x2+4x+4) = 2(x+2)(x+2) = 2(x+2)2—again

a perfect � !Exercise 6.8.


6.9. Answers:(a) x2 − 9 = (x − 3)(x + 3).

(b) x2 − 12 = (x − √12)(x +

√12) = (x − 2

√3)(x + 2

√3). You did

simplify, didn’t you?(c) x2 − 17 = (x − √

17)(x +√

17).(d) 25 − x2 = (5 − x)(5 + x) = −(x − 5)(x + 5). We usually like the

x-term to come first.

These factorizations are rather boring. Exercise 6.9.


6.10. Answers:

(a) 4x2 − 9 = (2x − 3)(2x + 3).

(b) 5x2 − 3 = (√

5x − √3)(

√5x +

√3).

(c) x4 − 25 = (x2 − 5)(x2 + 5) = (x − √5)(x +

√5)(x2 + 5).

Exercise 6.10.


6.11. Solutions:

(a) x2 − (x− 1)2 = [x− (x− 1)][x+(x− 1)] = [1][2x− 1] = 2x − 1

(b) x2y2 − 4 = (xy − 2)(xy + 2)

(c) Factor:(2x − 1)2 − (2x + 1)2

x.

(2x − 1)2 − (2x + 1)2

x

=[(2x − 1) − (2x + 1)][(2x − 1) + (2x + 1)]

x

=[−2][4x]

x= −8x

x� cancel the x!

= −8

That simplified down nicely!Exercise 6.11.


6.12. Solution:

8x2 + 2x − 1 = (4x − 1)(2x + 1).

Note: You really don’t need the answers, just multiply out yourfactorization. If it expands to 8x2 + 2x − 1, you are right!

Exercise 6.12.


6.13. Solution:

6x2 − 5x − 6 = (3x + 2)(2x − 3)

Exercise 6.13.


6.14. Answers:

(a) x3 − 1 = (x − 1)(x2 + x + 1)

(b) 27x3 − 8 = (3x − 2)(9x2 + 6x + 4) . (b = 3x and a = 2)

(c) 8y6 + 27 = (2y2 + 3)(4y4 − 6y2 + 9) (b = 2y2 and a = 3)Exercise 6.14.


6.15. Answers:

(a)x3 − 1x − 1

=(x − 1)(x2 + x + 1)

x − 1= x2 + x + 1

(b) Factor and simplify:(8x3 − 1)(x2 + 3x + 2)

x2 − 1.

(8x3 − 1)(x2 + 3x + 2)x2 − 1

=(2x − 1)(4x2 + 2x + 1)(x + 1)(x + 2)

(x − 1)(x + 1)

=(2x − 1)(4x2 + 2x + 1)(x + 2)

(x − 1)

(c)8x3 − 274x2 − 9

=2x − 3)(4x2 + 6x + 9)

(2x − 3)(2x + 3)=

4x2 + 6x + 92x + 3

Exercise 6.15.


6.1. Solution: The terms of the divisor, x − 1, and the dividend,2x3 − 3x2 + 6x − 1, are already arranged in decreasing order. (Step(a) of the Division Algorithm.

The calculations outlined in the Division Algorithm can be ar-ranged into a convenient table format. The method of division andthe table is similar to long division of numbers.

2x2− x + 5 � the quotient

x − 1 2x3−3x2+6x−4 � (b) divide 2x3 by x to get 2x2

2x3−2x2 � (c) multiply 2x2 by x − 1 and place it here

− x2+6x−4 � (c) subtract; (b) divide −x2 by x to get −x

− x2+ x � (c) multiply −x by x − 1 and place it here

5x−4 � (c) subtract; (b) divide 5x by x to get 5

5x−5 � (c) multiply 5 by x − 1 and place it here

1 � (c) subtract; Done. This is the remainder


Interpretation of Calculations: x − 1 goes into 2x3 − 3x2 + 6x − 4 . . .2x2 − x + 5 times with a remainder of 1. Thus,

2x3 − 3x2 + 6x − 4x − 1

= 2x2 − x + 5 +1

x − 1

Compare this equation with (2). The quotient is Q(x) = 2x2 − x + 5and the remainder is R(x) = 1. Note that deg(R(x)) = 0 < 1, asadvertised above.

The validity of the above equation can be verified by getting a commondenominator of the right-hand side of the equation and observing thatit results in the left-hand side! Example 6.1.


6.2. Problem (b) Divide:3x3 − 2x2 + 7x + 1

x2 + 3x − 2.

Solution to (b)

3x −11 � the quotient

x2 + 3x − 2 3x3− 2x2+ 7x+ 1 � (b) divide 3x3 by x2 to get 3x

3x3+ 9x2− 6x � (c) multiply 3x by x2 + 3x − 2

−11x2+13x+ 1 � (c) subtract; then repeat (b)

−11x2−33x+22 � (c) multiply −11 by x2 + 3x − 2

46x−21 � (c) subtract; Done. The remainder

Thus,3x3 − 2x2 + 7x + 1

x2 + 3x − 2= 3x − 11 +

46x − 21x2 + 3x − 2

Notice that the degree of the remainder is less than the degree of thedivisor.


Solution to (d) Divide:4x4 − 3x3 + 2x2 + 8x − 3

x + 3

4x3−15x2+ 47x− 133 � the quotient

x + 3 4x4− 3x3+ 2x2+ 8x− 3 � (b) divide

4x4+12x3 � (c) multiply

−15x3+ 2x2+ 8x− 3 � (c) subtract; (b) divide

−15x3−45x2 � (c) multiply

47x2+ 8x− 3 � (c) subtract; (b) divide

47x2+141x � (c) multiply

−133x− 3 � (c) subtract; (b) divide

−133x−399 � (c) multiply

396 � (c) subtract and Done.

Thus,

4x4 − 3x3 + 2x2 + 8x − 3x + 3

= 4x3 − 15x2 + 47x − 133 +396

x + 3

Example 6.2.


6.3. Solution: Here the denominator, the divisor, is a monomial : asingle term. In this case, we don’t have to carry out the elaboratedivision process. We simply . . .

4x12 − 7x9 + 1x3 =

4x12

x3 − 7x9

x3 +1x3

= 4x9 − 7x6 +1x3 .

That is, we divide by separation of fractions. Thus,

4x12 − 7x9 + 1x3 = 4x9 − 7x6 +

1x3

When a shorter method works . . . use the shorter method!Example 6.3.


6.4. Solutions:

(a) Factor x2 −x−2. Here we seek two numbers r1 and r2 such thatr1 + r2 = −1 and r1r2 = −1.Solution: List all pairs of integers the product of which is −2.

r1 r2 r1r2 r1 + r2−1 2 2 1−2 1 2 −1 ⇐= This is it

Therefore, the factorization is for r1 = −2 and r2 = 1:

x2 − x − 2 = (x + r1)(x + r2)

= (x + (−2))(x + 1)

= (x − 2)(x + 1).

Or,x2 − x − 2 = (x − 2)(x + 1).


(b) Factor x2 + 5x + 6.Solution: We follow the strategy suggested above. List all pairsof integers the product of which is 6, and we search for that pairwhose sum is 5.

r1 r2 r1r2 r1 + r21 6 6 7

−1 −6 6 −72 3 6 5 ⇐= This is it

Notice that I did not list out all possible combinations. I stoppedas soon as I found the proper pair: r1 = 2 and r2 = 3. Thefactorization is

x2 + 5x + 6 = (x + 2)(x + 3).


(c) Factor 4 − 3x − x2.Solution: This has a slight twist. The coefficient of the x2 termis not one. We just factor out −1 to get

4 − 3x − x2 = −(x2 + 3x − 4)

and factor x2 + 3x − 4. Again, we follow the strategy suggestedabove. List all pairs of integers the product of which is −4, andwe search for that pair whose sum is 3.

r1 r2 r1r2 r1 + r21 −4 −4 −3

−1 4 −4 3 ⇐= Found them!

The factorization is

4 − 3x − x2 = −(x2 + 3x − 4) = −(x − 1)(x + 4).

Or, we could, perhaps write is as follows:

4 − 3x − x2 = (1 − x)(x + 4).

Example 6.4.


6.5. Solution: This is not a pretty method. The problem is to factor6x2 + x − 1.

Trial 1: Guess a factorization of the form (6x+r1)(x+r1). The valuesof r1 and r2 are such that r1r2 = −1.

Try: r1 = 1, r2 = −1: (6x + 1)(x − 1) = 6x2 − 5x − 1 Error!

Try: r1 = −1, r2 = 1: (6x − 1)(x + 1) = 6x2 + 5x − 1 Error!

Trial 2: Guess a factorization of the form (3x + r1)(2x + r2). Thevalues of r1 and r2 are such that r1r2 = −1.

Try: r1 = 1, r2 = −1: (3x + 1)(2x − 1) = 6x2 − x − 1 Error!

Try: r1 = −1, r2 = 1: (3x − 1)(2x + 1) = 6x2 + x − 1 Success!

The factorization is,

6x2 + x − 1 = (3x − 1)(2x + 1).

Example 6.5.

Important Points


It is stated in the definition that a polynomial is reducible if it canwritten as a product of polynomials of smaller degree. In the factor-ization

x2 + 1 =12(2x2 + 2),

it is true that 12 has smaller degree than x2 + 1, but the other factor,

2x2 + 2, has degree equal to that of x2 + 1. Therefore, the abovefactorization does not mean x2 + 1 is reducible—it’s irreducible.

This is why the phrase “of lesser degree” was included in the definitionof reducible. Important Point


No. Simply factoring a polynomial by taking out a numerical constantsays nothing about whether that polynomial is reducible or not. Thisparticular polynomial 2x2 + 2x is reducible because

2x2 + 2x = 2x(x + 1).

We have factored a degree two polynomial into a product of two degree1 polynomials. Each factor has degree less than the original polyno-mial 2x2 + 2. Important Point

��mptiimenu


Lesson 7: Solving Equations & Inequalities

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 7: Solving Equations & Inequalities

Table of Contents7. Solving Equations & Inequalities7.1. How to Solve Linear Equations

• Essentially Linear Equations7.2. Solving Second Degree Equations

• Factoring Methods • Completing the Square • The Qua-dratic Formula

7.3. Solving Inequalities• Tools for Solving Inequalities • Simple Inequalities • Dou-ble Inequalities • The Method of Sign Charts

7.4. Solving Absolute Inequalities• Solving the Inequality |a| < b • Solving the Inequality|a| > b

7. Solving Equations & InequalitiesThere are two basic tools for solving equations: (1) Adding the sameexpression to both sides of an equation and (2) multiplying both sidesof the equation by the same expression. In symbols,

a = b ⇐⇒ a + c = b + c (1)

a = b ⇐⇒ ac = bc c �= 0 (2)

Here the symbol ⇐⇒ means “if and only if” which is a fancy way ofsaying “is equivalent to.” Equation (2) has an obvious variation

a = b ⇐⇒ a

c=

b

cc �= 0 (3)

Some care needs to be taken when applying (2) and (3) in the casewhere c is an algebraic expression containing unknowns. Usually, stu-dents have no problems when c is a numerical value.

Section 7: Solving Equations & Inequalities

7.1. How to Solve Linear Equations

A linear equation is an equation of the form

ax + b = 0 a �= 0

The solution set would represent the zeros or roots of the linear poly-nomial ax + b.

Even though the equation is very simple, student still make errorssolving it. Here are some representative examples. All linear equationsare solved in the same way.

Example 7.1. Solve each linear equation.(a) 4x + 5 = 0 (b) 1

2x − 4 = 6 (c) 7 − 3x = 2

Strategy for Solving. Isolate the unknown, x say, on one side of theequation with other terms of the equation on the other side. Dividethrough both sides by the coefficient of the unknown to obtain thesolution.

Exercise 7.1. Solve for x in each of the following using good tech-niques (as exhibited in Example 7.1). Passing is 100% correct.


(a) 4x − 8 = 0 (b) 23 − 2x = 3 (c)43x − 2 = 3 (d)

23x +

45=

83

Here is a slight variation on the type of linear equation previouslyconsidered.

Exercise 7.2. Solve each for x.(a) 3x − 2 = 7x + 3 (b) 9x + 3 = 4x − 2 (c)

12x =

43(x − 3)

Sometimes we have equations that have several symbolic quantities.Examine each of the following problems.

Exercise 7.3. In each of the equations listed below, solve for theindicated variable. Solve for . . .(a) x in 5x − 3y = 4 (b) y in 5x − 3y = 4(c) z in x2z − 12x + y = 1

It is apparent from the above examples and exercises that the opera-tions of adding (1), multiplying (2), and dividing (3) both sides of anequation by the same expression are standard and useful tools in your


toolbox of techniques for solving equations. Do not make up your ownmethods, use the standard ones and . . . these are they!

When working out mathematical problems it is important to organizeyour thoughts on paper properly and clearly. After solving a problem,recopy it neatly. Copy the style of this tutorial or some other textbook.Try to improve your handwriting. Use proper notation.

• Essentially Linear EquationsSome equations are ‘disguised’ linear equations. They do not requireany special techniques other than what is needed to solve linear equa-tions.

Example 7.2. Solve for x in each of the following.

(a) (x − 1)2 = x2 (b)2x2 + 5x − 1

x + 1= 2x + 1

(c)1 − 3x2x + 1

= 4 (d)4

3x + 1= 1


Now, consider the following set of exercises.

Exercise 7.4. Solve for x is each of the following.

(a)x

3x + 8= 5 (b)

5x + 23 − 8x

= 2

(c) (2x − 3)2 = (2x − 7)2 (d)8x2 + 2x + 1

2x + 1= 4x + 1

7.2. Solving Second Degree Equations

We now turn to the problem of solving equations of the form:

ax2 + bx + c = 0. (4)

There are two standard methods of solving this kind of equation:(1) by factoring the left-hand side and (2) by applying the BinomialFormula.


• Factoring MethodsWe have already studied techniques of factoring polynomials of de-gree two; therefore, it is not necessary to look at a large number ofexamples. If necessary, review factoring.

The method of factoring can certainly be applied to any polynomialequation and is not restricted to quadratic equations. In additionto factoring, a major tool used in solving equations is the Zero-Product Principle.

Zero-Product Principleab = 0 =⇒ a = 0 or b = 0 (5)

This principle states the obvious property of the real number system:The only way the product of two numbers can be zero is if one ofthem is zero.


The following example illustrates standard reasoning and solutionmethods. Read carefully.

Example 7.3. Solve each of the following.(a) x2 − 5x + 6 = 0 (b) x2 + 4x + 4 = 0 (c) 6x2 − x − 2 = 0

Now using the same methods as exhibited in the previous examplesolve each of the following. Passing is 100%.

Exercise 7.5. (Skill Level 1) Solve for x in each of the following.

(a) x2 − 7x + 12 = 0 (b) x2 + 3x = 10 (c)x2 + 5x − 6

x2 + 1= 0

Here are a few more quadratic equations.

Exercise 7.6. Solve for x in each of the following.

(a) 12x2 − 17x + 6 = 0 (b) 20x2 + 3x = 2 (c)4x2 − 4x + 1

x − 1= 0

Factorization techniques are not limited to second degree equations.Here are a few higher degree equations that can be factored fairly


easily. Some of the fourth degree equations below can be solved usingthe factorization techniques for quadratic polynomials.

Exercise 7.7. Solve for x in each of the following.(a) x3 − 2x2 − 3x = 0 (b) x4 − 16 = 0(c) x4 − 2x2 − 3 = 0 (d) x4 − 5x2 + 6 = 0

• Completing the SquareIn addition to factorization methods, the technique of completingthe square can also be used to solve a quadratic equation. Eventhough this technique will be used to solve equation, completion ofthe square has certain uses in other kinds of mathematical problems.If you go on to Calculus, for example, you will see it within thecontext of integration problem.

Completion of the Square Algorithm. Below are the steps for com-pleting the square with an abstract and a particular equation thatillustrate the steps.


1. Problem: Solve the quadratic equation for x:

ax2 + bx + c = 0 2x2 + 12x − 3 = 0.

2. Associate the x2 term and the x term:

(ax2 + bx) + c = 0 (2x2 + 12x) − 3 = 0

3. Factor out of the coefficient of the x2 term.

a(x2 +b

ax) + c = 0 2(x2 + 6x) − 3 = 0

4. Take one-half of the coefficient of the x and square it :

take the coefficient of x . . .b

a6

and compute one-half this . . .b

2a3

and square it :b2

4a2 9


5. Take this number, and add it inside the parentheses; this addi-tion must be compensated for by subtracting it from outside theparentheses:

a(x2 +b

ax+

b2

4a2 ) + c − b2

4a= 0 2(x2 + 6x+ 9)− 3− 18 = 0

Care must be made here because we are adding the term in-side the parentheses and subtracting an equal quantity outsidethe parentheses. Study the abstract version and the particularexample closely to understand what is meant.

6. The trinomial inside the parentheses is a perfect square:

a(x +b

2a)2 + c − b2

4a= 0 2(x + 3)2 − 21 = 0 (6)

End Complete Square

What does this accomplish? Observe that the equations in (6) can berewritten as

AX2 = C (7)


where X is a linear polynomial. This kind of equation can easily besolved as follows:

AX2 = C =⇒ X2 =C

A=⇒ X = ±

√C

A.

This sequence of steps can be carried out in every case.

Let’s illustrate by continuing to solve the equation carried in the com-pletion of the square algorithm.

Example 7.4. Solve the equation 2x2 + 12x − 3 = 0.

Example 7.5. Solve by completing the square: 3x2 + 2x − 5 = 0.

Exercise 7.8. Solve each of the following by completing the square.(a) 8x2 − 2x − 1 = 0 (b) 3x2 + 5x + 2 = 0 (c) x2 + x − 1 = 0

All the above examples and exercises were done exacly the same way.The method of completion of the square is a useful tool when solvingquadratic equations, but it is not the most efficient method. The nextsection on the Quadratic Formula is a more standard tool than iscompleting the square.


Despite its inefficiencies, completion of squares is still is useful tech-nique to know as it is use elsewhere in mathematics.

• The Quadratic FormulaThe solutions of equation (4) can be found in a more direct way thanthe method of factorization or completing the square by using theso-called Quadratic Formula. Let’s state/prove this formula.

Theorem. Consider the quadratic equation

ax2 + bx + c = 0 a �= 0 (8)(1) If b2 − 4ac < 0, (8) has no solutions;(2) if b2 − 4ac = 0, (8) has only one solution;(3) if b2 − 4ac > 0, (8) has two distinct solutions.

In the latter two cases, the solutions are given by the QuadraticFormula:

x =−b ± √

b2 − 4ac

2a(9)


Proof.Theorem Notes: The expression b2−4ac is called the discriminant forthe quadratic equation. It can be used, at casual glance, to determinewhether a given equation has one, two, or no solutions.

The discriminant is a handy way of classifying a polynomialP (x) = ax2 + bx + c as irreducible or not. The polynomial P (x) isirreducible if and only if its discriminant is negative: b2 − 4ac < 0.

Quiz. Using the discriminant, b2−4ac, respond to each of the followingquestions.1. Is the quadratic polynomial x2 − 4x + 3 irreducible?

(a) Yes (b) No2. Is the quadratic polynomial 2x2 − 4x + 3 irreducible?

(a) Yes (b) No3. How many solutions does the equation 2x2 − 3x − 2 = 0 have?

(a) none (b) one (c) twoEndQuiz.

Let’s go to the examples.


Example 7.6. Solve each of the following using the QuadraticFormula(a) x2 − 5x + 6 = 0 (b) x2 + 4x + 4 = 0(c) 6x2 − x − 2 = 0 (d) 3x2 − 3x + 1 = 0

Exercise 7.9. Using the quadratic formula, solve each of the follow-ing.(a) 2x2 + 5x − 12 = 0 (b) 3x2 − 7x + 1 = 0(c) x2 + 1 = 0 (d) x2 + x = 3

Recognition. One problem students have is recognizing a quadraticequation. This is especially true when the equation has several sym-bolic expressions in it. Basically, a quadratic equation in x is anequation in which x2 appears as a term and x appears as a term.The coefficient of the x2 term is the value of a; the coefficient of thex term is the value of b; and all other terms comprise the value of c.The symbol x may be some other letter like y or z, but it can also bea compound symbol like x2, y3, or even something like sin(x)!


Illustration 1. Here are some examples of quadratic equations.(a) The equation 3zx2−sx+4 = z is quadratic in x: a = 3z, b = −s

and c = 4− z. (Recall: We must write the equation in the form3zx2 − sx + (4 − z) = 0).)

(b) The equation 5wx4 − 2w2x2 + 3 = 0 is quadratic in x2 and alsoquadratic in w.

(c) The equation 4 sin2(x)−4 sin(x)+1 = 0 is a quadratic equationin the sin(x).

Comments: In example (b), the fact that the equation is quadraticin x2 means we can use the quadratic formula to solve for x2. Thesame equation is quadratic in w means that we can use the quadraticformula to solve for w. That’s the way it works!

Part of the power of the Quadratic Formula is that it is a veryefficient way of solving quadratic equations—more efficient than fac-toring, usually—and it can be applied even when the coefficients a, b,and c are symbolic. Here is a simple example.


Illustration 2. Solve for x in the equation yx2 − 2x− 4y2 = 0. Thisis a quadratic equation in x because the highest power of x is power2. We can apply the Quadratic Formula with a = y, b = −2 andc = −4y2:

x =−(−2) ± √

(−2)2 − 4(y)(−4y2)2y

=2 ±

√4 + 16y3

2y=

2 ± 2√1 + 4y3

2y

(In the last step, I’ve extracted 4 from the radical, which appears asa 2 outside the radical.)

Note that we only get a solution when 1+ 4y3 ≥ 0. This occurs wheny ≥ −1/ 3

√4. (Solving Inequalities is taken up later.)

Thus, for any y, y ≥ −1/ 3√4, the solutions for x are

x =1 +

√1 + 4y3

y, x =

1 −√1 + 4y3

y


Exercise 7.10. The equation that appears in Illustration 2 is aquadratic equation in y. Use the Quadratic Formula to solve fory.

Exercise 7.11. Use the Quadratic Formula to (a) solve for w inIllustration 1; and (b) solve for x2 in the same equation.

7.3. Solving Inequalities

We encounter the problem of solving inequalities in a variety of set-tings. For example, we stumbled across questions of solving inequali-ties in the previous section on the Quadratic Formula. In that section,our solutions came out in terms of symbolic quantities; recall the so-lution to the equation discussed in Illustration 2:

x =1 ±

√1 + 4y3

y.

We only have “real solutions” (as opposed to solutions that are com-plex numbers) provided

1 + 4y3 ≥ 0.


This is a (rather simple) problem of solving an inequality.

• Tools for Solving InequalitiesWhen you manipulate an inequality for the purpose of trying to isolatethe unknown on one side of the inequality, there are a few things youshould know.

Tools for Manipulating Inequalities: a ≤ b =⇒ a + c ≤ b + c (10)

a ≤ b and c > 0 =⇒ ac ≤ bc (11)

a ≤ b and c < 0 =⇒ ac ≥ bc (12)

a ≤ b and n ∈ N =⇒ n√

a ≤ n√

b (13)

Tool Notes: Equation (10) states that you can add the same quantityto both sides of an inequality, and the inequality is preserved.


By equation (11), you can multiply both sides of an inequalityby a positive number and the inequality will be preserved.

An important variation on this is equation (12): If you multiplyboth sides by a negative number the inequality is reversed. Caveat. Itis property (12) that most students have trouble with.

Finally, we can take a root of both sides and the inequality ispreserved. There is a built-in proviso: Provided the nth roots of bothsides exist. This is not a problem when n is odd, but becomes onewhen n is even.Example 7.7. Let’s finish the analysis of Illustration 2: In thesolution

x =1 ±

√1 + 4y3

y

we require 1 + 4y3 ≥ 0. Solve for y in this inequality.


• Simple InequalitiesThe basic tools can be used to solve a simple class of inequalities:Essentially Linear Inequalities.

Example 7.8. Solve each of the following inequalities for x.(a) 5x + 7 < 0 (b) 3 − 9x ≥ 4 (c) 3x5 + 4 ≥ 9 (d) 3x2 + 4 ≤ 3

Exercise 7.12. Solve each of the following inequalities for x. Writeyour solutions in interval notation.(a) 4x + 12 ≤ 5 (b) 1

2x − 32 > 4 (c) 3 − 8x ≥ 4

Here’s a slight variation on the previous problems.

Exercise 7.13. Solve each inequality for x. Leave your answer in setnotation.(a) 2x − 1 ≤ 5x + 2 (b) 3x + 4 > 1 − 6x

• Double InequalitiesBy double inequalities I mean inequalities of the form

a ≤ b ≤ c (14)


as well as all variations on same. This notation is short-hand for

a ≤ b and b ≤ c.

In this section we look at one simple type of problem—the case wherethe b in (14) is a linear polynomial and a and b are constants.

This kind of problem can be solve in much the same way as in theprevious paragraphs.

Illustration 3. Solve for x in −4 ≤ 3x − 1 < 6.Solution:

−4 ≤ 3x − 1 < 6 � given

−4 + 1 ≤ 3x < 6 + 1 � add 1 to all sides

−3 ≤ 3x < 7 � combine

−1 ≤ x < 73 � multiply all sides by 1/3

Presentation of Solution. Set Notation: { x | −1 ≤ x < 73 }


Interval Notation: [−1, 73 )

Illustration Notes: The standard tools for manipulating inequalitieshold. You can add the same quantity to all sides of the inequalitiesand you can multiply the same quantity to all sides.

If you multiply all sides by a negative number, all the inequalitiesare reversed!Exercise 7.14. Solve each of the following for x. Use interval no-tation to express your answer. Passing is 100%.(a) 3 < 2x + 1 ≤ 12 (b) −2 ≤ 2 − 5x ≤ −1 (c) 1 < 3

2x + 1 < 4

Quiz.1. What is the solution set to the inequality 1 ≤ x ≥ −1?

(a) [1,+∞) (b) [−1, 1] (c) ∅ (i.e., no solution)2. What is the solution to the inequality −1 ≥ x ≥ 1?

(a) [−1, 1] (b) [1,+∞) (c) ∅ (i.e., no solution)3. What is the solution to the inequality 3 ≤ 2x − 1 ≤ 1?

(a) [1, 2] (b) ∅ (i.e., no solution)EndQuiz.


• The Method of Sign ChartsMore complex inequalities require different methods. TheMethod ofSign Charts is a general method of analyzing inequalities providedyou can factor the algebraic expressions. Let me illustrate this methodwith an example.

Suppose you have an inequality of the form

R(x) ≥ 0

where R(x) is a rational expression. The idea is to factor the expres-sion R(x) completely—both numerator and denominator—and ana-lyze the sign of each factor. The Sign Chart is just a graphical wayof storing all the information.

The next example is important because it delineates the steps and thereasoning that goes into the Sign Chart Method.Read this examplecarefully.

Example 7.9. Solve the inequalities for x:(a) x2 − 3x + 2 ≥ 0 (b) x ≤ x2.


Exercise 7.15. Solve for x in each of the following.(a) x2 − x − 2 ≤ 0 (b) x3 − 4x2 + 3x > 0

Tip. The basis for the Sign Chart Method is a comparison of somealgebraic expression with zero:

R(x) < 0 or R(x) > 0 or R(x) ≤ 0 or R(x) ≥ 0

Therefore, the first step is to put your inequality in one of the aboveforms.

Exercise 7.16. Solve each of the following for x. Write your answersin interval notation.(a) x2 − 3x ≤ 4 (b) 6x − x2 < 5

This method is not limited to polynomial inequalities. The next ex-ample illustrates the method for an inequality involving a rationalexpression.

Example 7.10. Solve the inequality for x:x2 − 4x − 3

< 0.


Exercise 7.17. Solve for x in each of the following.

(a)x2 − 2xx + 1

< 0 (b)x

x2 − 3x + 2≥ 0

The Sign Chart Method is quite general and can be applied in a widevariety of problems. To illustrate this statement, the next exercise con-tinues Exercise 7.10, which, in turn, was problem originally startedin Illustration 2.

Exercise 7.18. In order to have real solutions to the equation inExercise 7.10, we require that the radicand in the solution

y =x2 ± √

x4 − 32x8

be nonnegative. Find all x for which x4 − 32x ≥ 0.

Let’s finish this section with a more challenging problem.


•Exercise 7.19. From Exercise 7.11, we used the quadratic formulato solve a fourth degree equation in x having symbolic coefficients interms of w. We obtained

x2 =w2 ± √

w4 − 15w5w

(15)

In order for there to be real solutions (solutions belonging to the realnumber system), we require (1) w4 − 15w ≥ 0 and (2) the right-handside of (15) to be nonnegative. Your assignment, should you decide toaccept it, find all values of w that satisfy conditions (1) and (2).

7.4. Solving Absolute Inequalities

We finish this lesson by a short study of inequalities that involve theabsolute value function. We look at two type of inequalities:

|a| < b and |a| > b.

These two are a fundamental type in inequality that occurs ratherfrequently in Calculus.


The method of solution is to (1) remove the absolute value . . . in alegal way; (2) solve the resultant inequality using standard techniquesdescribed previously.

• Solving the Inequality |a| < bThe key to removing the absolute value from |a| < b is in the inter-pretation of absolute value. Recall that

|a| = the distance a is from zero (0).

The inequality |a| < b then states that a is less than b units from theorigin. The fact that a is less than b units away from the origin wouldbe equivalent to saying that a is between −b and b. That is,

|a| < b is equivalent to − b < a < b

Let’s elevate this observation to the status of shadow box.


How to remove the | · | from |a| < b.

|a| < b ⇐⇒ −b < a < b (16)

Note. Naturally, a similar statement is true for |a| ≤ b; this is true ifand only if −b ≤ a ≤ b.

Once the absolute value is removed, we now have a double inequality—which we solve.

Example 7.11. Solve each of the following for x.(a) |x − 4| < 3 (b) |3x − 1| < 2 (c) |2 − 4x| ≤ 5

The method is simple enough, remove the absolute value, then solvethe double linear inequality.

Exercise 7.20. Solve each of the following for x. Leave your answerin interval notation. Passing is 100%.(a) |x + 3| < 8 (b) |4x + 9| ≤ 1 (c) |2 − 7x| ≤ 3


• Solving the Inequality |a| > bThe solution to the inequality |a| > b can be concluded once theabsolute value as been removed.

A number a satisfies the inequality |a| > b if the distance it is awayfrom the origin is greater than b units. This means the value of a isnot in the interval [−b, b ] (for these are exactly the numbers that arewithin b units of the origin). The number a we are looking for is not inthe interval [−b, b ]; therefore, the number a we seek must be greaterthan b or less than −b. In symbols . . .

How to remove the | · | from |a| > b.

|a| > b ⇐⇒ a > b or a < −b (17)

Note. A similar statement is true for |a| ≥ b: |a| ≥ b if and only ifa ≥ b or a ≤ −b.


The Split-Solve-Join Solution Method. The process of solving absoluteinequalities requires three steps.

1. Split: Split the absolute inequality using display line (17),2. Solve: Solve the two inequalities, obtaining their solution sets.3. Join: Join (take the union of) the solution sets from Step 2 to

obtain the final solution set to your absolute inequality.

Here are a couple of examples that illustrate the SSJS Method. Readcarefully.

Example 7.12. Solve each of the following for x.(a) |x − 3| > 4 (b) |5x + 1| ≥ 3

Exercise 7.21. Using the SSJS Method, solve each of the absoluteinequalities for x. Write your answer in interval notation.(a) |9x − 2| ≥ 3 (b) |2 − 3x| > 6 (c) | 32x + 2| > 1

3

These same methods can be applied to more complicated absoluteinequalities. Here we have just considered absolute linear inequalities.


Equations and inequalities are the way we can pose questions; there-fore, it is essential that we have very solid methods of solving equationsand inequalities. The techniques are demonstrated in this lesson arefar from being comprehensive, yet they will suffice you as you continueto learn.

In the next lesson, we introduce the Cartesian Coordinate Sys-tem and discuss Functions. In Lesson 9 you will see how we useequations and inequalities to pose questions; the techniques of thislesson are then applied to solve.

Click here to continue to Lesson 8.


7.1. Answers:

(a) 4x − 8 = 0 =⇒ 4x = 8 =⇒ x = 2

(b) 23 − 2x = 3 =⇒ −2x = −20 =⇒ x = 10

(c)43x − 2 = 3 =⇒ 4

3x = 5 =⇒ x =

34(5) =⇒ x =

154

Part (d) on the next page.


(d)23x +

45=

83.

Solution: The most difficult thing about this one is the arith-metic!

23x +

45=

83

� given

23x =

83

− 45

� subtract 4/5 both sides

23x =

2815

� arithmetic

x =32

· 2815

� multiply both sides by 3/2

x =145

Presentation of Answer : x =145

Exercise 7.1.


7.2. Solutions:

(a) Solve for x: 3x − 2 = 7x + 3.3x − 2 = 7x + 3 � given

3x = 7x + 5 � add 2 to both sides

−4x = 5 � add −7x to both sides

x = −54

� divide by −4

Presentation of Answer : x = −54

(b) Solve for x: 9x + 3 = 4x − 2.9x + 3 = 4x − 2 � given

9x = 4x − 5 � add −3 to both sides

5x = −5 � add −4x to both sides

x = −1 � divide both sides by −1



(c) Solve for x:12x =

43(x − 3).

12x =

43(x − 3) � given

3x = 8(x − 3) � multiply both sides by 6

3x = 8x − 24 � expand

−5x = −24 � add −8x to both sides

x =245

� divide both sides by 5


Exercise 7.2.


7.3. Answers: Make sure you understand the method of solution.The answers are given only here.(a) Solve for x in 5x − 3y = 4.

x =3y + 4

5

(b) Solve for y in 5x − 3y = 4.

y =5x − 4

3

(c) Solve for z in x2z − 12x + y = 1.

z =1 + 12x − y

x2

Exercise 7.3.


7.4. Solutions:

(a) Solvex

3x + 8= 5.

x

3x + 8= 5 � given

x = 5(3x + 8) � mulitply both sides by 3x + 8

x = 15x + 40 � expand

−40 = 14x � add −x − 40 to both sides

14x = −40 � transpose

x = −4014

� divide by 14

x = −207

� reduce fractions


.


(b) Solve5x + 23 − 8x

= 2.

5x + 23 − 8x

= 2 � given

5x + 2 = 2(3 − 8x) � multiply both sides by 3 − 8x

5x + 2 = 6 − 16x � expand r.h.s.

5x + 16x = 6 − 2 � add 16x − 2 to both sides

21x = 4 � combine similar terms

x =421

� divide by 21


.


(c) Solve (2x − 3)2 = (2x − 7)2.

(2x − 3)2 = (2x − 7)2 � given

4x2 − 12x + 9 = 4x2 − 28x + 49 � expand using (5), Lesson 5

−12x + 9 = −28x + 49 � add −4x2 to both sides

28x − 12x = 49 − 9 � add 28x − 9 to both sides

16x = 40 � combine

x =4016

� divide by 16

x =52

Presentation of Answer : x =52.

Comment : This problem is similar to (a) of Example 7.2; however, inmy solution I gave a more “traditional’ solution. In the second lineabove, I simply expanded the binomials—this is perhaps what youdid yourself. The rest follows using standard methods.


(d) Solve8x2 + 2x + 1

2x + 1= 4x + 1.

8x2 + 2x + 12x + 1

= 4x + 1 � given

8x2 + 2x + 1 = (2x + 1)(4x + 1) � multiply both sides by 2x + 1

8x2 + 2x + 1 = 8x2 + 6x + 1 � expand using (2) of Lesson 5

2x + 1 = 6x + 1 � add −8x2 to both sides

−4x = 0 � add −6x − 1 to both sides

x = 0 � divide both sides by −4

Presentation of Answer : x = 0

Now what do you think of that! Exercise 7.4.


7.5. Solutions:(a) Solve for x: x2 − 7x + 12 = 0.

x2 − 7x + 12 = 0

(x − 3)(x − 4) = 0therefore, either

x − 3 = 0x = 3

or x − 4 = 0x = 4

Presentation of Solution: x = 3, 4(b) Solve for x: x2 + 3x = 10.

x2 + 3x = 10

x2 + 3x − 10 = 0

(x + 5)(x − 2) = 0therefore, either

x + 5 = 0x = −5

or x − 2 = 0x = 2

Presentation of Solution: x = −5, 2


(c) Solve for x:x2 + 5x − 6

x2 + 1= 0.

x2 + 5x − 6x2 + 1

= 0

x2 + 5x − 6 = 0 � mulitply both sides by x2 + 1

(x + 6)(x − 1) = 0

therefore, either

x + 6 = 0x = −6

or x − 1 = 0x = 1

Presentation of Answer : x = −6, 1Exercise 7.5.


7.6. Answers:(a) 12x2 − 17x + 6 = 0

12x2 − 17x + 6 = 0 � given

(3x − 2)(4x − 3) = 0 � factor it!

From this we can see that the solutions are

3x − 2 = 0 or 4x − 3 = 0

x =23

x =34

Presentation of Solutions: x =23,34

(b) 20x2 + 3x = 2.

20x2 + 3x − 2 = 0 � add −2 to both sides

(5x + 2)(4x − 1) = 0 � factor it!


Thus,5x + 2 = 0 or 4x − 1 = 0

x = −25

x =14

Presentation of Solutions: x = −25,14

(c)4x2 − 4x + 1

x − 1= 0.

4x2 − 4x + 1 = 0 � multiply both sides by x − 1

(2x − 1)2 = 0 � factor it–perfect squrare!

Presentation of Solution: x =12

Exercise 7.6.


7.7. Answers:

(a) x3 − 2x2 − 3x = 0. Factoring this we obtain

x(x2 − 2x − 3) = 0

x(x − 3)(x + 1) = 0

Presentation of Solutions: x = 0, 3,−1

(b) x4 − 16 = 0. Let’s factor—difference of two squares!

(x2 − 4)(x2 + 4) = 0

(x − 2)(x + 2)(x2 + 4) = 0

Presentation of Solutions: x = −2, 2

Comments: The last factor x2+4 is an irreducible quadratic—itcannot be factored.

(c) x4 − 2x2 − 3 = 0. This is a quadratic equation in the variablex2: (x2)2 − 2(x2)− 3 = 0. If you don’t understand what I mean,temporarily put y = x2; our equation becomes y2 − 2y − 3 = 0.


This is clearly a quadratic in y, but y = x2, so it is a quadraticin x2. Let’s factor it using the factoring techniques.

(x2 − 3)(x2 + 1) = 0 � factor!

(x −√3)(x +

√3)(x2 + 1) = 0 � again!

My the Zero-Product Prinicple, we then have

x −√3 = 0 or x +

√3 = 0

x =√3 x = −

√3

Presentation of Solutions: x = −√3,

√3

(d) x4 − 5x2 + 6 = 0. This is again quadratic in x2.

(x2 − 3)(x2 − 2) = 0 � factor!

(x −√3)(x +

√3)(x −

√2)(x +

√2) = 0 � again!


By the Zero-Product Prinicple, we then have

x −√3 = 0 =⇒ x =

√3

or,x +

√3 = 0 =⇒ x = −

√3

or,x −

√2 = 0 =⇒ x =

√2

or,x +

√2 = 0 =⇒ x = −

√2

Presentation of Solutions: x = −√3,

√3,−

√2,

√2Exercise 7.7.


7.8. Solutions:(a) Solve for x: 8x2 − 2x − 1 = 0.

8x2 − 2x − 1 = 0 � given

8(x2 − 14x) = 1 � Steps 2 & 3

8(x2 − 12x + 1

64 ) = 1 + 864 � Step 4

8(x − 18 )

2 = 98 � perfect square

(x − 18 )

2 = 964 � divide by 8

x − 18 = ± 3

8 � take square root

x = 18 ± 3

8 � add 1/8 to both sides

Presentation of Solutions: x = − 14 , 1

2

Comments: Here, my solution uses a slight variation in the tech-niques illustrated in the examples. Rather than having all termon the left-hand side, I took the constant term to the right-hand


side, then when I completed the square, I added 864 to both sides

of the equation.(b) Solve for x: 3x2 + 5x − 2 = 0.

3x2 + 5x + 2 = 0 � given

3(x2 + 53x) = −2 � Steps 2 & 3

3(x2 + 53x + 25

36 ) = −2 + 3( 2536 ) � Step 4

3(x + 56 )


(x + 56 )

2 = 136 � divide by 8

x + 56 = ± 1


x = − 56 ± 1

6 � add 1/8 to both sides

Presentation of Solution: x = − 23 ,−1


(c) Solve for x: x2 + x − 1 = 0.

x2 + x − 1 = 0 � given

x2 + x = 1 � Steps 2 & 3

(x2 + x + 14 ) = 1 + 1

4 � Step 4

(x + 12 )


x + 12 = ±

√5


x = − 12 ±

√5

2 � add −1/2 to both sides

Presentation of Solution: x = − 12 +

√5

2 ,− 12 −

√5

2 or,

Presentation of Solution: x =√5 − 12

,−√5 + 12

(Verify!)

Exercise 7.8.


7.9. Solutions:(a) Solve for x: 2x2 + 5x − 12 = 0.

x =−5 ± √

52 − 4(2)(−12)2(2)

� a = 2, b = 5, c = −12

=−5 ± √

1214

=−5 ± 11

4

=32, −4

Presentation of Solution: x =32,−4


(b) Solve for x: 3x2 − 7x + 1 = 0

x =−(−7) ± √

(−7)2 − 4(3)(1)2(3)

� a = 3, b = −7, c = 1

=7 +

√37

6,7 − √

376

Presentation of Solutions: x =7 +

√37

6,7 − √

376

(c) Solve for x: x2 + 1 = 0

b2 − 4ac = 02 − 4(1)(1) = −4 < 0 � a = 1, b = 0, c = 1

Therefore, this equation has no solutions.(d) Solve for x: x2+x = 3. Begin by putting it into the proper form:

x2 + x − 3 = 0.

x =−1 ± √

1 − 4(1)(−3)2

� a = 1, b = 1, c = −3

=−1 ± √

132


Presentation of Solutions: x =−1 +

√13

2,

−1 − √13

2Exercise 7.9.


7.10. Solution: Solve for y in yx2 − 2x − 4y2 = 0. The first thing todo is to rearrange the equation:

−4y2 + x2y − 2x = 0

Now we can see that a = −4, b = x2, and c = −2x.

y =−x2 ± √

(x2)2 − 4(−4)(−2x)2(−4)

=−x2 ± √

x4 − 32x(−8)

= −−x2 ± √x4 − 32x8

=x2 ± √

x4 − 32x8

Presentation of Solutions:

y =x2 +

√x4 − 32x8

,x2 − √

x4 − 32x8

Comments: We get real solutions when x is such that x4 − 32x ≥ 0.This problem will be continued in the section on Solving Inequalities.

Exercise 7.10.


7.11. Solution to (a): Begin by rearranging the equation in the formof a quadratic in w:

5wx4 − 2w2x2 + 3 = 0becomes

(−2x2)w2 + (5x4)w + 3

Now it is apparent that a = −2x2, b = 5x4, and c = 3.

w =−5x4 ± √

(5x4)2 − 4(−2x2)(3)2(−2x2)

= −−5x4 ± √25x8 + 24x2

4x2

= −−5x4 ± |x|√25x6 + 244x2

Now here’s an interesting simplification. How to remove the absolutevalue? If x ≥ 0, the |x| = x; if x < 0, then |x| = −x. The second casehas the effect of changing ‘±’ (plus or minus) to ‘∓’ (minus or plus).


So I think it is safe to remove the absolute value—it will not changethe solutions. Therefore,

w = −−5x4 ± x√25x6 + 24

4x2

= −−5x3 ± √25x6 + 24

4xNow x is a factor common to both the numerator and denominator—we can and do cancel it.

Presentation of Answer : w = −−5x3 ± √25x6 + 24

4x

Did you get it?


Solution to (b): Solve for x2 in 5wx4 − 2w2x2 + 3 = 0.

x2 =−(−2w2) ± √

(−2w2)2 − 4(5w)(3)2(5w)

=2w2 ± √

4w4 − 60w10w

=2w2 ± 2

√w4 − 15w

10w

=w2 ± √

w4 − 15w5w

Presentation of Solution:

x2 =w2 ± √

w4 − 15w5w

(A-1)

Comments: Real solutions exist provided w4 − 15w ≥ 0. When wetake up solving inequalities, we’ll return to this problem to do a more


detailed analysis of it. Suffice it to say that solutions exist for x2

provided (1) w4 − 15w ≥ 0 and (2) the right-hand side of (A-1) mustbe nonnegative as well.

Exact conditions under which this equation has solution for x2 can bemade after we analyze some inequalities . . . later. Exercise 7.11.


7.12. Solutions:(a) Solve for x: 4x + 12 ≤ 5.

4x + 12 ≤ 5 � given

4x ≤ −7 � add −12 to both sides (10)

x ≤ −74

� multiply both sides by 1/4: see (11)

Presentation of Answer. Interval Notation:(

−∞,−74,]

(b) Solve for x: 12x − 3

2 > 4.

12x − 3

2 > 4 � given

x − 3 > 8 � multiply both sides by 2: (11)

x > 11 � add 3 to both sides: see: (10)

Presentation of Answer. Interval Notation: ( 11,+∞ )


(c) Solve for x: 3 − 8x ≥ 4

3 − 8x ≥ 4 � given

−8x ≥ −1 � add −3 to both sides: (10)

x ≤ 18

� inequality reversed! See: (12)


−∞,18

]Exercise 7.12.


7.13. Solutons:(a) Solve for x: 2x − 1 ≤ 5x + 2.

2x − 1 ≤ 5x + 2 � given

−3x ≤ 3 � add 1 − 5x to both sides

x ≥ −1 � inequality reversed: (12)

Presentation of Answer : { x | x ≥ −1 }(b) Solve for x: 3x + 4 > 1 − 6x.

3x + 4 > 1 − 6x � given

9x > −3 � add 6x − 4 to both sides: (10)

x > −13

� inequalities not reversed: (10)

Presentation of Answer :{

x | x > −13

}Exercise 7.13.


7.14. Solutions:(a) Solve for x: 3 < 2x + 1 ≤ 12.

3 < 2x + 1 ≤ 12 � given

2 < 2x ≤ 11 � add −1 to all sides

1 < x ≤ 112

� multiply 1/2 on all sides

Presentation of Answer :(1,

112

](b) Solve for x: −2 ≤ 2 − 5x ≤ −1.

−2 ≤ 2 − 5x ≤ −1 � given

−4 ≤ −5x ≤ −3 � add −2 to all sides

45

≥ x ≥ 35

Presentation of Answer :[45,35

]


(c) Solve for x: 1 < 32x + 1 < 4.

1 < 32x + 1 < 4 � given

2 < 3x + 2 < 8 � multiply all sides by 2

0 < 3x < 6 � add −2 to all sides

0 < x < 2 � multiply all sides by 1/3

Presentation of Answer : ( 0, 2 )Exercise 7.14.


7.15. Solutions to(a): Solve for x: x2 −x−2 ≤ 0. Begin by factoringthe quadratic.

(x + 1)(x − 2) ≤ 0

Now, do a Sign Chart Analysis.The Sign Chart of (x + 1)(x − 2)

-1x + 1

2x − 2

-1 2(x + 1)(x − 2)

legend : • negative (−)• positive (+)

Presentation of Solution. Interval Notation: [−1, 2 ]

Set Notation: { x | −1 ≤ x ≤ 2 }

Comments: We include the endpoints in the solution because theywould satisfy the inequality—that’s a good reason!


Solution to (b): Solve for x: x3 − 4x2 + 3x > 0. Begin by factoringcompletely:

x(x − 1)(x − 3) > 0.

The Sign Chart of x(x − 1)(x − 3)

0x

1x − 1

3x − 3

0 1 3x(x − 1)(x − 3)


Presentation of Solution. Interval Notation: ( 0, 1 ) ∪ ( 3,+∞ )

Set Notation: { x | 0 < x < 1 or x > 3 }


Comments: Here, we do not include the endpoints. At the endpoints,the expression is zero, which does not satisfy the inequality.

Exercise 7.15.


7.16. Solution to (a): Solve for x in x2 − 3x ≤ 4. Begin by takingthe 4 to the left-hand side of the inequality,

x2 − 3x − 4 ≤ 0.

Now factor,(x + 1)(x − 4) ≤ 0.

Now do a Sign Chart for (x + 1)(x − 4).

The Sign Chart of (x + 1)(x − 4)

-1x + 1

4x − 4

-1 4(x + 1)(x − 4)


Presentation of Solution: [−1, 4 ]


Solution to (b): Solve for x in 6x − x2 < 5. Therefore,

6x − x2 < 5 � given

6x − x2 − 5 < 0 � put everything to l.h.s.

x2 − 6x + 5 > 0 � multiply by −1: I didn’t like the −x2

(x − 1)(x − 5) > 0 � now factor!

Now do a Sign Chart for (x − 1)(x − 5).

The Sign Chart of (x − 1)(x − 5)

1x − 1

5x − 5

1 5(x − 1)(x − 5)


Presentation of Solution: (−∞, 1, ] ∪ [ 5,+∞ ) Exercise 7.16.


7.17. Solution to (a): Solvex2 − 2xx + 1

< 0.

Begin by factoring:x(x − 2)

x + 1< 0. Now do a Sign Chart.

The Sign Chart ofx(x − 2)(x + 1)

0x

2x − 2

-1x + 1

-1 0 2

x(x − 2)x + 1



Presentation of Solution: Solvex(x − 2)

x + 1< 0.

Set Notation: { x | x < −1 or 0 < x < 2 }Interval Notation: (−∞,−1 ) ∪ ( 0, 2 )

Solution to (b) is on the next page.


Solution to (b): Solve for x:x

x2 − 3x + 2≥ 0.

Begin by factoring,x

(x − 1)(x − 2)≥ 0

It is this inequality we solve using the Sign Chart Method.

The Sign Chart ofx

(x − 1)(x − 2)

0x

1x − 1

2x − 2

0 1 2

x

(x − 1)(x − 2)



Presentation of Solution: Solvex

(x − 1)(x − 2)≥ 0.

Set Notation: { x | 0 ≤ x < 1 or x > 2 }Interval Notation: [ 0, 1 ) ∪ ( 2,+∞ )

We include x = 0 in the solution because this point makes the expres-sion 0; hence, x = 0 satisfies the inequality. We don’t include x = 1 orx = 2 in the solution set because they make the denominator equalto zero—a no-no. Exercise 7.17.


7.18. Solution: Solve x4 − 32x ≥ 0. First factor: x(x3 − 32) ≥ 0.

The second factor can be factored some more (it is the difference incubes), but in this instance it would be a waste of pencil lead to do so.The important point about the factors is that you can discern whenthey are positive and when they are negative—we can do that for thefactor (x3 − 32).

We must ask the question, when is x3 − 32 > 0? The Answer: whenx3 > 32 or when x > 3

√32. Similarly, x3 − 32 < 0 when x < 3

√32.

Thus, . . .

The Sign Chart of x(x3 − 32)

0x

3√32x3 − 32

0 3√32x(x3 − 32)



Presentation of Solution: Solve for x: x4 − 32x ≥ 0.

Interval Notation: (−∞, 0 ] ∪ [ 3√32,+∞ )

Comments: This finally answers the question: For what values of xdoes the equation

yx2 − 2x − 4y2 = 0

have real solutions for y? The answer is that x must be a numberless than or equal to 0 or a number greater than or equal to 3

√32.

Otherwise, there are no real solutions for y. Exercise 7.18.


7.19. Solution: Using the Method of Signs we can easily solve theinequality w4 − 15w ≥ 0 in exactly the same way we did in Exer-cise 7.18. Thus,

w4 − 15w ≥ 0 has solution w ≤ 0 or w ≥ 3√15

To continue the analysis, we need to break our argument down intocases.

Case 1. w ≥ 3√15, in particular w is positive.

Since w > 0, w4 − 15w < w4. Take the square root of both sides weget

√w4 − 15w < w2. This implies that

w2 ±√

w4 − 15w > 0

since we are adding/subtracting a smaller number. But this means,

x2 =w2 ± √

w4 − 15w5w

> 0

since both numerator and denominator are positive.


Case 2. w ≤ 0. Actually, w �= 0 since w appears in the denominator.

Since w < 0, w4 < w4−15w, since we are subtracting a negative num-ber, the result will be larger. Taking roots, we get w2 <

√w4 − 15w.

Therefore,

w2 +√

w4 − 15w > 0 but w2 −√

w4 − 15w < 0

This means,

w2 +√

w4 − 15w5w

< 0 andw2 − √

w4 − 15w5w

> 0

since, you’ll recall, we are assuming in this case that w < 0.

Summary.1. For any w ≥ 3

√15,

w2 ± √w4 − 15w5w

> 0.


In this case, there are four solutions for x.

x2 =w2 ± √

w4 − 15w5w

=⇒ x = ±√

w2 ± √w4 − 15w5w

Ouch!2. For any w < 0, only

w2 − √w4 − 15w5w

> 0.

In this case, there are two solutions for x.

x2 =w2 − √

w4 − 15w5w

=⇒ x = ±√

w2 − √w4 − 15w5w

That was ugly! Exercise 7.19.


7.20. Solutions:(a) Solve for x: |x + 3| < 8.

|x + 3| < 8 � given

−8 < x + 3 < 8 � from (16)

−11 < x < 5 � add −3 to all sides

Presentation of Solution: (−11, 5 )

(b) Solve for x: |4x + 9| ≤ 1.

|4x + 9| ≤ 1 � given

−1 ≤ 4x + 9 ≤ 1 � from (16)

−10 ≤ 4x ≤ −8 � add −9 to all sides

− 104 ≤ x ≤ − 8

4 � multiply all sides by 1/4

Now, reducing to lowest terms we get the . . .

Presentation of Solution: [− 52 ,−2 ]


(c) Solve for x: |2 − 7x| ≤ 3.

|2 − 7x| ≤ 3 � given

−3 ≤ 2 − 7x ≤ 3 � from (16)

−5 ≤ −7x ≤ 1 � add −2 to all sides

57 ≥ x ≥ − 1

7 � multiply all sides by −1/7or

− 17 ≤ x ≤ 5

7

In the last step we have multiplied both sides by a negativenumber, this will reverse the direction of the inequality !

Presentation of Solution: [− 17 , 5

7 ]Exercise 7.20.


7.21. Solution to (a) Solve for x: |9x − 2| ≥ 3.

|9x − 2| ≥ 3

Use (17) to split the inequality!

9x − 2 ≥ 3 � upper inequality

9x ≥ 5 � add 2 both sides

x ≥ 59 � divide by 9

[ 59 ,+∞) � solution set

9x − 2 ≤ −3 � lower inequality

9x ≤ −1 � add 2 both sides

x ≤ − 19 � divide by 9

(−∞,− 19 ] � solution set

Now, join the solutions!

Solution Set = [59 ,+∞) ∪ (−∞,− 19 ]

Presentation of Solution: (−∞,− 19 ] ∪ [ 59 ,+∞)


Solution to (b) Solve for x: |2 − 3x| > 6.

|2 − 3x| > 6


2 − 3x > 6 � upper inequality

−3x > 4 � add −2

x < − 43 � divide by −3

(−∞,− 43 ) � solution set

2 − 3x < −6 � lower inequality

−3x < −8 � add −2

x > 83 � divide by 5

( 83 ,+∞) � solution set


Solution Set = (−∞,− 43 ) ∪ ( 8

3 ,+∞)

Presentation of Solution: (−∞,− 43 ) ∪ ( 8

3 ,+∞)

Comment : Hopefully, you understand when to include the endpointsand when not to include them in your solution set, and, most impor-tantly, how do denote the inclusion/exclusion of the endpoints.


Solution to (c) Solve for x: | 32x + 2| > 13 .

| 32x + 2| > 13


32x + 2 > 1

3 � upper inequality

9x + 12 > 2 � multiply by 6

9x > −10 � add −12

x > − 109 � divide by 9

(− 109 ,+∞) � solution set

32x + 2 < − 1

3 � lower inequality

9x + 12 < −2 � multiply by 6

9x < −14 � add −12

x < − 149 � divide by 9

(−∞,− 149 ) � solution set


Solution Set = (− 109 ,+∞) ∪ (−∞,− 14

9 )

Presentation of Solution: (−∞,− 149 ) ∪ (− 10

9 ,+∞) Exercise 7.21.


7.1.(a) Solve for x: 4x + 5 = 0

Solution:4x + 5 = 0 � given

(4x + 5) − 5 = −5 � subtract −5 from both sides

4x = −5 � combine

14(4x) =

14(−5) � multi. both sides by 1/4

x = −54

� combine and done!

Solving an equation consists of a sequence of legal steps: addingthe same quantity to both sides (see equation (1)); multiplyingboth sides by the same quantity (see equation (2)) were usedabove.


Presentation of Answer : x = −54.

(b) Solve for x: 12x − 4 = 6.

Solution:12x − 4 = 6 � given

(12x − 4) + 4 = 6 + 4 � add 4 to both sides

12x = 10 � combine

2(12x) = 2(10) � multi. both sides by 2

x = 20 � combine and done!

Note this equation initially was not of the form ax + b = 0.There was a nonzero constant on right-hand side of the equation.Obviously, the equation 1

2x−4 = 6 is equivalent to 12x−10 = 0;

consequently, no big deal was made of it.

Presentation of Solution: x = 20.


(c) Solve for x: 7 − 3x = 2.Solution:

7 − 3x = 2 � given

−(7 − 3x) = −2 � multiply both sides by −1

3x − 7 = −2 � simplify

(3x − 7) + 7 = −2 + 7 � add 7 to both sides

3x = 5 � combine

13(3x) =

13(5) � multiply both sides by 1/3

x =53

� combine and done!

Comments: Here the coefficient of x was negative. I opted tomultiply by −1 to ‘change the sign’ so that I would feel morecomfortable solving.

Presentation of Answer : x =53.

Example 7.1.


7.2. Solutions:

(a) Solve for x: (x − 1)2 = x2.

(x − 1)2 = x2� given

(x − 1)2 − x2 = 0 � sub. x2 both sides: see (1)

[(x − 1) − x][(x − 1) + x] = 0 � diff. of squares!

−(2x − 1) = 0 � combine

2x − 1 = 0 � multipy by −1: see (2)

2x = 1 � add 1 to both sides: see (1)

x =12

� divide by 2: see (3)

Comments: Note the recommended way of handling an equationwith squares on both sides. Rather than taking the square rootof both sides, take everything to one side of the equation andfactor it as a difference of squares: this is a much nicer method.The other option is to expand all binomials and combine.



(b) Solve for x:2x2 + 5x − 1

x + 1= 2x + 1

2x2 + 5x − 1x + 1

= 2x + 1 � given equation

2x2 + 5x − 1 = (x + 1)(2x + 1) � multiply both sides by x + 1

2x2 + 5x − 1 = 2x2 + 3x + 1 � expand r.h.s.

5x − 1 = 3x + 1 � sub. 2x2 from both sides: (1)

2x = 2 � two steps in one!

x = 1 � divide both sides by 2

Presentation of Answer : x = 1


(c) Solve for x:1 − 3x2x + 1

= 4.

1 − 3x2x + 1

= 4 � given equation

1 − 3x = 4(2x + 1) � multiply both sides by 2x + 1

1 − 3x = 8x + 4 � expand r.h.s.

−11x = 3 � add −8x − 1 to both sides

x = − 311

� divide both sides by 11

Presentation of Answer : x = − 311


(d) Solve for x:4

3x + 1= 1.

43x + 1

= 1 � the given equation

4 = 3x + 1 � multiply both sides by 3x + 1

3x + 1 = 4 � transpose equation: a = b ⇐⇒ b = a

3x = 3 � add −1 to both sides

x = 1 � divide both sides by 3

Presentation of Answer : x = 1Example 7.2.


7.3. Solutions:(a) Solve x2 − 5x + 6 = 0.

Solution:x2 − 5x + 6 = 0 � given

(x − 2)(x − 3) = 0 � factor l.h.s. (S-1)

From (S-1) and the Zero-Product Principle, we deduce thateither

x − 2 = 0 or x − 3 = 0x = 2 x = 3

Presentation of Solution: x = 2, 3 .

Comments: Note that this equation has two distinct solu-tions. An alternate method of presenting the solutions to anequation is to present the solution set: the set of all solutions.Here we could have written,

Solution Set = { 2, 3 }


(b) Solve x2 + 4x + 4 = 0.Solution:

x2 + 4x + 4 = 0 � given equation

(x + 2)2 = 0 � factor: perfect square!Therefore,

(x + 2) = 0 � Zero-Product Principleor,

x = −2


Comments: The left-hand side is a quadratic polynomial that,it turned out, was a perfect square. Note that this equation hasonly one solution.

(c) Solve 6x2 − x − 2 = 0.Solution: Let’s take the polynomial aside and factor it using thefactoring techniques explained in Lesson 6.


Factor: 6x2 − x − 2 = 0. Begin by trying a factorization of theform

6x2 − x − 2 = (3x + r1)(2x + r2),

where r1 and r2 are chosen so that r1r2 = −2 and the “cross-product” is correct. After some trial and error we get r1 = −2and r2 = 1; thus,

6x2 − x − 2 = (3x − 2)(2x + 1). (S-2)

Solve: Now let’s take the factorization (S-2) and return to ouroriginal equation:

(3x − 2)(2x + 1) = 0

We can now obtain the solutions by applying the Zero-ProductPrinciple: (3x − 2)(2x + 1) = 0 implies either . . .

3x − 2 = 0 or 2x + 1 = 03x = 2 2x = −1

x =23

x = −12


We deduce the solution set is { 23 ,− 1

2 }.

Presentation of Solution: x =23,−1

2Example 7.3.


7.4. Solution: This will be short. All the work has been already done.

2x2 + 12x − 3 = 0 � given

2(x + 3)2 − 21 = 0 � complete the square

2(x + 3)2 = 21 � equation of the form (7)

(x + 3)2 =212

� divide by 2

x + 3 = ±√

212

� take square root

x = −3 ±√

212

� solved!

Presentation of Solution: x = −3 +

√212

,−3 −√

212

Example 7.4.


7.5. Solution: Just follow the algorithm.

3x2 + 2x − 5 = 0 � given

3(x2 + 23x) − 5 = 0 � Step 2 & 3

3(x2 + 23x + 1

9 ) − 5 − 13 = 0 � Step 4: add/sub 1/2 coeff. of x

3(x + 13 )

2 = 163 � perfect square!

(x + 13 )

2 = 169 � divide by 3

x + 13 = ± 4

3 � take square root both sides

x = − 13 ± 4

3 � solve for x and done

Thus,

x = − 13 ± 4

3

= − 13 + 4

3 , − 13 − 4

3

= 1, − 53

Presentation of Solution: x = − 53 , 1 Example 7.5.


7.6. Solutions: We simply apply the Quadratic Formula!(a) Solve for x: x2 − 5x + 6 = 0. This is the case where a = 1,

b = −5, and c = 6:

x =−(−5) ± √

(−5)2 − 4(1)(6)2(1)

=5 ± √

25 − 242

=5 ± 12

=5 + 12

,5 − 12

=62,42

= 3, 2

Presentation of Answers: x = 2, 3


(b) Solve for x: x2+4x+4 = 0. This is the case where a = 1, b = 4,and c = 4.

x =−4 ± √

42 − 4(1)(4)2(1)

=−4 ± √

16 − 162

=−42

= −2

Presentation of Solution: x = −2

Comments: This is the case of the discriminant of zero. There-fore we have only one solution.


(c) Solve for x: 6x2 − x − 2 = 0. Here a = 6, b = −1, and c = −2.

x =−(−1) ± √

(−1)2 − 4(6)(−2)2(6)

=1 ± √

1 + 4812

=1 ± 712

=1 + 712

,1 − 712

=812

,−612

=23, −1

2

Presentation of Solutions: x =23,−1

2


(d) Solve for x: 3x2 − 3x + 1 = 0. This is the case of a = 3, b = −3and c = 1. Before starting the standard calculations, we mightcheck the discriminant:

b2 − 4ac = (−3)2 − 4(3)(1) = 9 − 12 = −3 < 0.

We’ve saved ourselves some work. This equation has no solu-tions.

Example 7.6.


7.7. Solve 1 + 4y3 ≥ 0. We simply use the tools:

1 + 4y3 ≥ 0 � given

4y3 ≥ −1 � add −1 both sides: see (10)

y3 ≥ −14

� multiply 1/4 both sides: see (11)

3√

y3 ≥ 3

√−14

� by (13)

y ≥ − 13√4

� apply properties of radicals: see (1)

The answer can be presented in three ways.

1. Inequalities. Solution: y ≥ − 13√4.

2. Interval Notation. Solution:[− 1

3√4,∞ )

.

3. Set Notation. Solution:{

y | y ≥ − 13√4

}.

Example 7.7.


7.8. Solutions:(a) Solve for x: 5x + 7 < 0.

x + 7 < 0 � given

5x < −7 � add −7 to both sides (10)

x < −75

� multiply both sides by 1/5: see (11)


−∞,−75

)(b) Solve for x 3 − 9x ≥ 4.

3 − 9x ≥ 4 � given

−9x ≥ 1 � add −3 to both sides

x ≤ −19

� mul. by −1/9, inequality reversed! See (12)

Presentation of Answer. Set Notation:{

x | x ≤ −19

}


(c) Solve for x: 3x5 + 4 ≥ 9.

3x5 + 4 ≥ 9 � given

3x5 ≥ 5 � add −4 to both sides

x5 ≥ 53

� multiply by 1/3

x ≥ 5

√53

� take 5th root both sides: See (1)

Presentation of Answer. Inequalities. x ≥ 5

√53

(d) Solve for x: 3x2 + 4 ≤ 3.

3x2 + 4 ≤ 3 � given

3x2 ≤ −1 � add −4 to both sides

x2 ≤ −13

� divide by 3


Now it is apparent that there are no solutions to this equation.For any x, x2 ≥ 0 so x2 cannot be less than a negative number.

Presentation of Solution. Set Notation. The solution set is ∅,the empty set.

Example 7.8.


7.9. Solution to (a): This method begins by factoring the algebraicexpressions:

(x − 1)(x − 2) ≥ 0 (S-3)

The idea is to analyze when each factor is positive and when eachfactor is negative. The Sign Chart is a graphical scheme for storingall the information. Here’s the Sign Chart for (x − 1)(x − 2):

The Sign Chart of (x − 1)(x − 2)

1x − 1

2x − 2

1 2(x − 1)(x − 2)


Sign Chart of x − 1. We ask the question, where is x − 1 > 0?Answer: x > 1. We mark that in blue on the axis. (When using penciland paper, I write little ‘+’ signs over the axis.) We now ask the


question, where is x − 1 < 0? Answer: x < 1. We mark that in red onthe axis. (Pencil and paper? Write little ‘−’ signs over the axis.) And,of course, x − 1 = 0 when x = 1. This represents a complete analysisof the factor x − 1: when its positive, when its negative, and when itszero.

Sign Chart of x − 2. We do the same analysis for the factor x − 2indicating the results in blue (or a ‘+’ ) or in red (or a ‘−’ ).

Sign Chart of (x − 1)(x − 2). To obtain the final sign chart forthe expression of interest, we put the other charts together using thefollowing principles:

(−)(−) = (+), (+)(−) = (−) and (+)(+) = (+)

For example, if we consider a number x < 1, then, according to mysign charts, the first factor (x − 1) is negative and the second factor(x − 2) is negative too; the product (x − 1)(x − 2) is positive and weindicate that on the sign chart for (x − 1)(x − 2). Get the idea?


The last axis tells us exactly when (x−1)(x−2) is positive and whenit is negative. We can see that the solution to the inequality

x2 − 3x + 2 ≥ 0

is all x in the blue. Members of the solution set are all x ≤ 1 plus allx ≥ 2. (We include x = 1 and x = 2 because they make the expressionx2 − 3x + 2 = 0 and would therefore satisfy the inequality.)


Set Notation: { x | x ≤ 1 or x ≥ 2 }Interval Notation: (−∞, 1 ] ∪ [ 2,+∞ )

Solution to: (b) Not as much detail will be given in this solution.

Problem: Solve for x in x < x2. The first thing we must do is totake all quantities to one side of the inequality: x − x2 < 0. Next, wemust factor, x(1 − x) < 0. Finally, we apply our Sign Chart Method.


The Sign Chart of x(1 − x)

0x

11 − x

0 1x(1 − x)


In the (1 − x) sign chart, we asked the question, when is 1 − x > 0.The answer is when 1 > x, or x < 1. This is depicted in blue.

Presentation of Solution: Solve x(1 − x) < 0.Set Notation: { x | x < 0 or x > 1 }Interval Notation: (−∞, 0 ) ∪ ( 1,+∞ )

Here, we exclude the endpoints, 0 and 1, because they do not satisfythe stated inequality. (These points make x(1 − x) equal zero.)

Example 7.9.


7.10. Solution: Begin by completely factoring the expression.

(x − 2)(x + 2)(x − 3)

< 0 (S-4)

We now do a Sign Chart on the left-hand side of (S-4).

The Sign Chart of(x − 2)(x + 2)

(x − 3)

-2x + 2

2x − 2

3x − 3

-2 2 3

(x + 2)(x − 2)x − 3



The solution set is all real numbers that are in red.

Presentation of Solution: (−∞,−2) ∪ (2, 3) Example 7.10.


7.11. Solution:(a) Solve for x: |x − 4| < 3.

|x − 4| < 3 � given

−3 < x − 4 < 3 � from (16)

1 < x < 7 � add 4 to all sides

Presentation of Solution: ( 1, 7 )

(b) Solve for x: |3x − 1| < 2.

|3x − 1| < 2 � given

−2 < 3x − 1 < 2 � from (16)

−1 < 3x < 3 � add 1 to all sides

− 13 < x < 1 � multiply all sides by 1/3

Presentation of Solution: (− 13 , 1 )


(c) Solve for x: |2 − 4x| ≤ 5.

|2 − 4x| ≤ 5 � given

−5 ≤ 2 − 4x ≤ 5 � from (16)

−7 ≤ −4x ≤ 3 � add −2 to all sides

74 ≥ x ≥ − 3

4 � multiply all sides by −1/4or

− 34 ≤ x ≤ 7

4

In the last step we have multiplied both sides by a negativenumber, this will reverse the direction of the inequality !

Presentation of Solution: [− 34 , 7

4 ]Example 7.11.


7.12. Solution to (a): Solve for x: |x − 3| > 4.

|x − 3| > 4


x − 3 > 4 � upper inequality

x > 7 � add 3 both sides

(7,+∞) � solution set

x − 3 < −4 � lower inequality

x < −1 � add 3 both sides

(−∞,−1) � solution set


Solution Set = (7,+∞) ∪ (−∞,−1)

Presentation of Solution: (−∞,−1) ∪ (7,+∞)

Comments That’s way it goes: split, solve each, and join by union!


Solution to (b): Solve for x: |5x + 1| ≥ 3.

|5x + 1| ≥ 3


5x + 1 ≥ 3 � upper inequality

5x ≥ 2 � add −1 both sides

x ≥ 25 � divide by 5

[ 25 ,+∞) � solution set

5x + 1 ≤ −3 � lower inequality

5x ≤ −4 � add −1 both sides

x ≤ − 45 � divide by 5

(−∞,− 45 ] � solution set


Solution Set = [25 ,+∞) ∪ (−∞,− 45 ]

Presentation of Solution: (−∞,− 45 ] ∪ [ 25 ,+∞) Example 7.12.

Important Points


Proof: The Quadratic Formula. The quadratic formula is nothingmore than the formula you get when you complete the square andsolve for x just as we did in the section on completing the square. Ifyou’ve been reading along this tutorial, the following steps would lookfamiliar.

ax2 + bx + c = 0 � given

ax2 + bx = −c � add −c to both sides

a(x2 +b

ax) = −c � factor out a from l.h.s.

a(x2 +b

ax +

b2

4a2 ) =b2

4a− c � add b2/4a to both sides

a(x +b

2a)2 =

b2 − 4ac

4a� perfect square

(x +b

2a)2 =

b2 − 4ac

4a2 � divide by a

All the steps above are reversible. This means that the solution set tothe last equation is the same as the solution set to the first equation.


You can see now that if b2 − 4ac < 0, the right-hand side of the lastequation is negative and the left-hand side is not; therefore, there canbe no solutions to the equation.

Now let’s continue the development.

x +b

2a= ±

√b2 − 4ac

4a2 � take square root of both sides

x +b

2a= ±

√b2 − 4ac

2a� extract root in denominator

x = − b

2a±

√b2 − 4ac

2a� subtract b/2a from both sides

x =−b ± √

b2 − 4ac

2a� combine fractions (I-1)

Now if b2 − 4ac = 0, then equation (I-1) becomes x = − b

2a; that is,

there is only one solution as asserted.


Finally, if b2 − 4ac > 0, then√

b2 − 4ac > 0 and formula (I-1) wouldclearly lead to two distinct solutions. Important Point


The solution to 1 ≤ x ≥ −1? The answer is [ 1,+∞). Sometimes yousee students write inequalities this way, but this is not a good way ofwriting a double inequality !

The question is, what does 1 ≤ x ≥ −1 mean? It means

1 ≤ x and x ≥ −1

Turn around the inequalities to get

x ≥ 1 and x ≥ −1

Now we ask ourselves the question: What are all the numbers x thatare greater than (or equal to) 1 and greater than (or equal to) −1?But any x satisfying x ≥ 1 automatically satisfies x ≥ −1; therefore,the important condition is x ≥ 1, hence the solution set in intervalnotation is [ 1,+∞ )

Students who write inequalities like 1 ≤ x ≥ −1 often mean x ≥ 1or x ≥ −1 and they just meld the two sets of inequalities together.However, the double inequality is for pairs of inequalities connected


logically by an “and”. This is different from pairs of inequalities con-nected by a logical “or”.

Make a distinction in your mind between the two: “and” �= “or” inmathematics or in English. Important Point


I often see students write this kind of inequality. The answer is theempty solution: ∅. Recall the meaning of double inequalities:

−1 ≥ x ≥ 1 means − 1 ≥ x and x ≥ 1or

x ≤ −1 and x ≥ 1

Now what number, x, is there that is less than or equal to −1 andgreater than or equal to 1? The Answer: There is no such numberthat satisfies both inequalities.

What the student actually means is x ≤ −1 or x ≥ 1, but they donot write it correctly.

Learn to use the notation to express precisely what you mean.Important Point


Again, the answer is no solution: ∅. If you went through the (rou-tine) steps of solving, you would have arrived at 2 ≤ x ≤ 1. Whatnumber x satisfies x ≥ 2 and x ≤ 1? Answer: No number.

However, you need not have bothered to solve an inequality havingthe empty solution set. It is immediate from the original inequality:

3 ≤ 2x − 1 ≤ 1

Look at the left-most and right-most expressions. Do you see that

3 ≤ 1?

This implies immediately that there are no solutions.Important Point

��mptiimenu


Lesson 8: Cartesian Coordinate System & Functions

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10

−(ab − (3ab − 4)) = 2ab − 4(ab)3(a−1 + b−1) = (ab)2(a+ b)(a − b)3 = a3 − 3a2b+ 3ab2 − b3

2x2 − 3x − 2 = (2x+ 1)(x − 2)12x+ 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }

f(x) = mx+ by = sinx

inTen Lessons



Lesson 8: Cartesian Coordinate System & Functions

Table of Contents8. Cartesian Coordinate System & Functions8.1. The Cartesian Coordinate System

• Referencing/Plotting Points • The Distance Formula • TheMidpoint of a Line Segment

8.2. Functions• The Definition • The Domain of a Function • Points ofIntersection of Curves

8. Cartesian Coordinate System & Functions

8.1. The Cartesian Coordinate System

There are many schemes for referencing points in the plane. Amongall these coordinate systems, the Cartesian Coordinate System isthe most popular and useful.

Figure 1

Begin by drawing two real number lines, perpendicular toeach other, and intersecting at their zeros. One number line isdrawn horizontally, and the other vertically.

The two number lines, called axes, are labeled by some appropriatelychosen symbols; usually the horizontal axis is labeled by the letter xand the vertical axis is labeled by the letter y. The horizontal axisis called the x-axis or the axis of abscissas and the vertical axis iscalled the y-axis or the axis of ordinates.

Figure 2

The two perpendicular axes subdivide the plane into six sub-sets in such a way that any given point in the plane is either(1) on the x-axis, (2) on the y-axis, (3) in the first quadrant,

Section 8: Cartesian Coordinate System & Functions

(4) in the second quadrant, (5) in the third quadrant, or (6) in thefourth quadrant. (See Figure 2.)

• Referencing/Plotting PointsPoints in the plane are referenced by their position relative to thetwo perpendicular axes. We shall not spend a terribly large amountof time on this topic because you no doubt have plotted many pointsbefore. We shall be content with reviewing some of the definitions andterminology.

Figure 3

The Method of Referencing a Point. Let P be a point in theplane. Draw a vertical line passing through the point P anda horizontal line through P . The vertical line intersects the

x-axis at a certain position a and the horizontal line intersects they-axis at a certain position b. The Cartesian coordinates of thepoint P is defined as

P ( a, b )

Conversely, given ordered pair of numbers, ( a, b ), there correspondsone and only one point in the plane. This point is the intersection of


the two lines obtained by drawing a vertical line passing through a inthe x-axis and a horizontal line passing through the number b on they-axis.

Terminology : Let ( a, b ) is the Cartesian Coordinates of the pointP . The number a is called the first coordinate of P , or the x-coordinate of P , or the abscissa of P ; similarly, b is called the secondcoordinate, or the y-coordinate, or the ordinate of P .

• Question. Why do you think we have such terminology as the “axisof abscissas,” the “axis of ordinates,” the “abscissa of P” and the“ordinate of P”?

Having defined the method of referencing a point in the plane, thefour quadrants of the plane can be described more precisely.

Quiz. Answer each of the following about the quadrants.1. What quadrant consists of all points P (x, y ) satisfying x > 0

and y < 0? Quadrant . . .

(a) I (b) II (c) III (d) IV


2. What quadrant consists of all points P (x, y ) satisfying x < 0and y > 0? Quadrant . . .

(a) I (b) II (c) III (d) IV3. What quadrant consists of all points P (x, y ) satisfying x < 0

and y < 0? Quadrant . . .

(a) I (b) II (c) III (d) IV4. What quadrant consists of all points P (x, y ) satisfying x > 0

and y > 0? Quadrant . . .

(a) I (b) II (c) III (d) IVEndQuiz.

• The Distance FormulaIn the next few paragraphs we take up the problem of computing thedistance between two points in the plane.


The Distance Formula:

Figure 4

Let P (x1, y1 ) and Q(x2, y2 ) be two points in theplane, then the distance between P and Q, denotedby d(P,Q ), is given by

d(P,Q ) =√(x1 − x2)2 + (y1 − y2)2 (1)

The validity of equation (1) is based on the Pythagorean Theo-rem. From Figure 4, we have

[d(P,Q )]2 = |x1 − x2|2 + |y1 − y2|2= (x1 − x2)2 + (y1 − y2)2,

whence comes equation (1).

Here is a typical ‘beginners’ example that illustrates how to use thisformula.


Example 8.1. Calculate the distance between the points P (−2, 4)and Q(3,−1).

Exercise 8.1. (Skill Level 0) In each of the follow, find the distance,d(P,Q ), between P and Q. Passing is 100%.(a) P (0, 0) and Q(3,−4) (b) P (−3, 4) and Q(−1,−1)(c) P (−4, 2) and Q(5, 2) (d) P (−1, 5) and Q(7, 9)

With the distance formula firmly in hand, we can now solve a numberof problems related to distances in the plane.

Exercise 8.2. Below, we define three points P , Q, and R. Determinewhether these three points are the vertices of a right triangle. (Hint :A triangle is a right triangle if and only if the square of the length ofthe longest side is equal to the sum of the squares of the other twosides.)(a) P (0, 0), Q(1, 1) and R(2, 0)(b) P (6,−7), Q(11,−3) and R(2,−2)(c) P (1, 2), Q(−3, 4) and R(4,−2)


The distance formula can also be used to test whether one point isbetween two others.

Quiz. Let P , Q, and R be three points in the plane. Which of thefollowing is equivalent to the statement that Q lies “between” P andR; i.e., Q lies on the straight line segment that connects P and R.(Hint : Think about the geometry of each equation—draw a picture ifnecessary.)(a) [d(P,R)]2 = [d(P,Q)]2 + [d(Q,R)]2

(b) d(P,R) = d(P,Q) + d(Q,R)(c) [d(P,Q)]2 = [d(P,R)]2 + [d(R,Q)]2

(d) d(P,Q) = d(P,R) + d(R,Q)

Exercise 8.3. Using the criterion stated in the above Quiz, deter-mine whether Q is between P and R.(a) P (1, 3), Q(2, 5) and R(4, 9)(b) P (−1, 10), Q(2,−5), and R(5,−12)

Exercise 8.4. The three points P (1,−1), Q(5, 0) and R(3, 1) are thevertices of a triangle. Compute the perimeter of the triangle PQR.


Exercise 8.5. A particle moves around in the xy-plane. It is knownthat at any given time t, the particle has coordinates P (1+t, 3−t). Atwhat time, t, will the particle be 4 units away from the origin, O(0, 0)?(Hint : Set up the equation d(P,O) = 4, and solve for t.)

Let us now look at a couple of special cases. Should you go on to Cal-culus, occasionally you will need to be able to quickly and efficientlycompute the distance between two horizontal points and the distancebetween two vertical points. In these two cases, it is not necessary touse the distance formula in its full generality. Read on.

� Distance between two Horizontally Oriented Points. Let’s begin witha quick quiz, the answer to which represents criterion for judgingwhether two points are horizontally oriented.

Quiz. Let P (x1, y1 ) and Q(x2, y2 ) be points in the plane. The P andQ are horizontally oriented is equivalent to the condition(a) x1 = x2 (b) y1 = y2


Now, let us developed a specialized version the distance formula. LetP (x1, y1 ) and Q(x2, y2 ) be two horizontally oriented points: Then,

d(P,Q ) =√(x1 − x2)2 + (y1 − y2)2

=√(x1 − x2)2 � since y1 = y2

= |x1 − x2 | � by (1) of Lesson 2

Distance Between Horizontally Oriented Points: Let P (x1, y1 ) andQ(x2, y2 ) be two points horizontally oriented points in the plane, i.e.y1 = y2, then the distance between P and Q, denoted by d(P,Q ), isgiven by

d(P,Q) = |x1 − x2| (2)

Comment : Usually, equation (2) is evaluated by taking the abscissaof the right-most point and subtracting the abscissa of the left-mostpoint.

Learn to use this specialized formula by answering the questions inthe quiz that follows.


Quiz. Work the solutions out first, then choose the correct response.Passing is 100%.1. Are the points (2, 6) and (−4, 6) horizontally oriented?

(a) Yes (b) No2. Which of the following is the distance between P (3,−9) and

Q(8,−9)?(a) −11 (b) −5 (c) 5 (d) 11

3. Let P (6,−3) and Q(−3,−3) be given. The value of d(P,Q) is(a) −9 (b) −3 (c) 3 (d) 9

4. Let P (−5, 2) and Q(−9, 2) be given. The value of d(Q,P ) is(a) 4 (b) 5 (c) 9 (d) 14

EndQuiz.

� Distance between two Vertically Oriented Points. Let P (x1, y1 ) andQ(x2, y2 ) be two points in the plane. Then P and Q are verticallyoriented if and only if x1 = x2, that is, if they have the same firstcoordinate.


The Distance Between Vertically Oriented Points: Let P (x1, y1 ) andQ(x2, y2 ) be two points vertically oriented points in the plane, i.e.x1 = x2, then the distance between P and Q, denoted by d(P,Q ), isgiven by

d(P,Q) = |y1 − y2| (3)

Comments: The distance between two vertically oriented points is theheight of the upper point minus the height of the lower point.

Exercise 8.6. Use the distance formula to derive equation (3).

Exercise 8.7. Calculate the distance between each of the followingsets of points and observe whether each pair of points horizontally orvertically oriented.(a) P (1, 2) and Q(1, 9) (b) P (−3, 3) and Q(−3,−4)(c) P (4, 3) and Q(−3, 3) (d) P (π,−5) and Q(π,−2)


• The Midpoint of a Line SegmentLet P (x1, y1) and Q(x2, y2) be two points in the plane. Draw a linesegment connecting the point P and Q.

Problem. Calculate the Cartesian coordinates of the midpoint ofthe line segment PQ.

Figure 5

Let us approach this problem as follows1: The coordinate xmust be halfway between x1 and x2; the coordinate y must behalfway between y1 and y2. Once we accept this reasoning, we

can deducex =

x1 + x2

2y =

y1 + y2

2.

(Note: We have used here the midpoint formula for the real numberline.) See Figure 5.

1Here is a brief discussion of an alternate approach.


Thus the Cartesian coordinates of the midpoint are

M

(x1 + x2

2,y1 + y2

2

)

Let’s elevate this formula to the status of a shadow box.

Midpoint Formula:Let P (x1, y1 ) and Q(x2, y2 ) be points in the plane. Thecoordinates of the midpoint M between P and Q are givenby

M

(x1 + x2

2,y1 + y2

2

)(4)

Exercise 8.8. (Skill Level 0) Calculate the midpoint between eachof the following pairs of points.(a) P (−1, 3) and Q(5, 7) (b) P (2, 4) and Q(2,−5)(c) P (5,−3) and Q(12,−3) (d) P (−1,−1) and Q(4, 2)


Exercise 8.9. Each pair of points listed below is diametrically op-posite to each other on a circle. Find the center, C(h, k), and radius,r, of the circle.(a) P (1, 2) and Q(−3,−1) (b) P (−1, 4) and Q(0, 0)

8.2. Functions

In the world of Mathematics one of the most common creatures en-countered is the function. It is important to understand the idea of afunction if you want to gain a thorough understanding of Algebra andCalculus.

Science concerns itself with the discovery of physical or scientific truth.In a portion of these investigations, researchers (or engineers) attemptto discern relationships between physical quantities of interest. Thereare many ways of interpreting the meaning of the word “relation-ships,” but in these lessons we are most often concerned with func-tional relationships. Roughly speaking, a functional relationship be-tween two variables is a relationship such that one of the two variables


has the property that knowledge of it (or knowledge of its value) im-plies a knowledge of the value of the other variable.

For example, the physical quantity of area, A, of a circle is relatedto the radius of that circle, r. Indeed, it is internationally knownthat A = πr2—an equation, I’m sure, you have had more than oneoccasion to examine in the past. The simple equation A = πr2 setsforth the principle of a functional relationship: Given knowledge of thevalue of one variable (the independent variable), r, then we have totalknowledge of the value of the other variable (the dependent variable),A. This causal (or deterministic) relationship one variable has withanother variable is the essence of a functional relationship.

This only difference between the example of the previous paragraphand any other example of a function, either one taken from the appliedfields or one that is of a more “purely abstract” nature, is the way inwhich the functional relationship is defined, and the complexity of thatdefinition. There are many, many ways of defining (or describing) afunctional relationship between one variable (or a set of variables) andanother variable (or another set of variables). Some of these methods


are rather “natural,” which you will encounter as you continue withthese lessons; others are “unnatural,” but we will not encounter themat this level of play.

Before we continue with this discussion, perhaps it is best to have aformalized definition of a function—in the next section.

• The Definition

Definition. Let A be a set and B be a set. A function, f , from A intoB is a rule that associates with each element in the set A a uniquecorresponding element in the set B. In this case, we write symbolically,f :A → B, or A

f→ B.

Definition Notes: The set A is called the domain of the function f .Typically in Algebra and Calculus, the set A will be an interval of thereal number line R. As a notation, we shall refer to the domain of thefunction f by Dom(f).

The set B is called the codomain of f . The set B may not bethe range of the function.


Let the elements of A are referred to by the letter x, and those ofthe set B by y. The symbol x is called the independent variable off , and y is called the dependent variable. The independent variablecan take on any value in the domain, Dom(f), of f .

A function f is a rule that associates with each element, x, inthe set A, a unique corresponding element, y, in the set B. The usualway we define a function is by an equation that states the relationshipbetween the variable x and the variable y.

For example, let the function f associate the number x with thenumber y, where y = x2. We write,

f :x → y where y = x2

or, more simplyf :x → x2

thus,f : 2 → 4 f :−3 → 9 f : 0 → 0

Where, f : 2 → 4 states that f associates with x = 2 the uniquecorresponding number y = 4.


The above notation is used frequently at higher levels of math-ematics, at our level, we use the standard functional notation. Ratherthan writing f :x → x2, we simply write f(x) = x2.

f(x) = x2 means f :x → x2.

Particular evaluations are carried out as follows:

f(2) = 22 = 4 f(−3) = (−3)2 = 9 f(0) = 02 = 0

More examples are given below.Given a particular x in Dom(f), y = f(x) is called a value of

the function f . For the function f(x) = x2, since f(2) = 4, we cansay that 4 (y = 4) is a value of f . It should be clear to you that −4in not a value of the function f(x) = x2.

For any given function, f , some numbers are values of f whileothers are not. The set of all values of a given function is called itsrange, denoted by Rng(f); thus,

Rng(f) = { y | y is a value of f }


To be a value, a ‘y’ must be of the form f(x) for some x. Thus,

Rng(f) = { y | y = f(x) for some x in Dom(f) }For the function f(x) = x2, the range is

Rng(f) = [ 0,+∞ )

Do you understand why?All the functions that we encounter in Algebra (and most of thoseencountered in Calculus) defined using algebraic expression in oneunknown. For example, the expression x2 − 4x + 1 is an algebraicexpression in x. We can use this expression to define a function by

f(x) = x2 − 4x+ 1

Here are a few examples of functions defined this way. This method isby no means the only way of defining functions. Read these examplescompletely and carefully.

Illustration 1. Examples of functions and numerical evaluations.


(a) Define f by f(x) = x2 + x. Then,

f(3) = 32 + 3 = 12 f(−3) = (−3)2 + (−3) = 6.

(b) Define g by g(x) =x

x+ 1. Then,

g(2) =2

2 + 1=

23

g(− 12 ) =

12

12 + 1

=1232

=13.

The names of functions are determined by the user, that’s youand me. I choose a name of g this time.

(c) Define h by h(t) =√t. Then,

h(4) =√4 = 2 h(9) = 3 h(5) =

√5.

Now I have changed the letter used to denote the independentvariable. I have used t instead of the traditional x—this causesno problems I hope? Any letter (or symbol) can be used for theindependent variable, and any letter (or symbol) can be usedfor the dependent variable.


(d) Any symbol you say? How about defining a function W byW (φ) = φ3. Then,

W (−5) = −125 W (3) = 27 W (√2 ) = (

√2 )3 = 2

√2.

Here, I have used the Greek letter φ (“phi”) for the name of theindependent variable.

Evaluation Tip. To evaluate a function, such as f(x) = x2 − 2x, ata particular value of x = −1, first replace the independent variablex with the number −1, then evaluate the expression. Thus, f(−1) =(−1)2 − 2(−1) (replace x by −1), then evaluate f(−1) = 1 + 2 = 3.Note the use of the parentheses: this is necessary because we arereplacing a single letter x by a compound symbol −1. Not to includethese parentheses (1) is mathematically and notationally wrong, and(2) invites evaluation errors.

Exercise 8.10. Evaluate each of the functions defined below at theindicated values. Passing is 100%.(a) f(x) = 2x2 − 3x; f(2), f(−2), f(− 1

2 )(b) g(s) = s(s+ 1)(s+ 2); g(0), g(1), g(−1), g(−3)


(c) h(t) =t

t2 + 1; h(1), h(−2), h( 1

2 ), h(− 12 )

(d) D(w) = w√w; D(9); D( 1

9 )

• For those Who want Greater Insight. Models for Functions.Listed behind this link is a description of several ways in which wecan view a function. These points of view may help you to understandthis important mathematical object.

• The Domain of a FunctionIn the examples in the previous paragraphs, nothing was mentionedconcerning the domains of the functions considered. In this sectionwe briefly discuss methods of computing the domain of a function.

The domain of the function defined is either (1) explicitly specified oris (2) not explicitly specified. (That seems reasonable.)

Illustration 2. Examples of functions with explicitly specified do-mains.


(a) Define a function f by

f(x) = x2, x ≥ 1

Here, we are explicitly defining the domain of f to be

Dom(f) = [ 1,+∞ ) = {x | x ≥ 1 }For this function, f(2) = 4 is defined, but f(0) is not becausex = 0 does not fall into the specified domain.

(b) Define a function g by

g(x) =x

x2 + 1, 0 ≤ x < 1

Here, we have specified the domain of g to be

Dom(g) = [ 0, 1 ) = {x | 0 ≤ x < 1 }Illustration Notes: Such (artificial) restriction of the domains mayarise from physical considerations. Perhaps these functions, f and gabove, are modeling some physical system; within the context of this


physical system, it only make sense to consider x ≥ 1, in the case ofthe function f , and 0 ≤ x < 1 in the case of g.When the domain of a function at definition time is left unspecified,that usually means we are to take as the domain the so-called naturaldomain of the function.

� The Natural Domain of a function. Given a function y = f(x). Thenatural domain of f is the set of all real numbers, x, for which thevalue f(x) can be calculated as a real number.

The next example illustrates the reasoning and methods used to cal-culate the natural domain of a function. Read carefully!

Example 8.2. Compute the natural domain of each of the following.

(a) f(x) = x2 + 3x+ 1 (b) g(x) =x2

x2 − 3x+ 2(c) h(x) =

√x+ 2 (d) p(x) =

√x2 − 1


Strategy. Let y = f(x) be a function, where f(x) is some algebraicexpression. The natural domain consists of all x for which . . .

• the denominator (if any) is not equal to zero; and• any radicands of even roots (if any) are nonnegative.

Here’s another example that incorporates all components of the abovestrategy.

Example 8.3. Find the natural domain of f(x) =√

x

x+ 1.

Quite typically, the strategy involves setting up some constraints orconditions on the values of the independent variable in the form ofinequalities. Once you identify these inequalities, you solve them (pos-sibly using the Sign Chart Method). The natural domain is then theset of all values of the independent variable that satisfy all the con-straints or conditions.

Exercise 8.11. Compute the natural domain of each of the following.

(a) f(x) = |x − 1| (b) g(x) =3x

x2 + 2x − 8


(c) h(x) =1√

x(x+ 2)(d) p(x) =

√x

x2 − 1

• Points of Intersection of CurvesWe have seen in the previous section that determining the naturaldomain of a functions oftimes require the setting up and solving ofinequalities. To determine where two curves intersect, if at all, we mustbe able to set up and solve equations. This is why the basic mechan-ics of solving inequalites and equations are so important—equationsand inequalities are the natural way in which we ask questions andthe techniques of solutions are the way we are able to answer thesequestions.

� Determining the x-intercept. Let y = f(x) be a function. The x-intercept, if there is one, is that value of x such that f(x) = 0. Asyou know, every function has a graph—graphing will be taken up inLesson 9—and in terms of the graph the x-intercept is the locationon the x-axis where the graph crosses the x-axis.


Procedure. To find the x-intercept(s) of the function y = f(x), set upthe equation

f(x) = 0 (5)

and solve for x.

Exercise 8.12. Find the x-intercept(s), if any, of each of the follow-ing functions.(a) f(x) = 4x − 1 (b) f(x) = x2 − 3x+ 2(c) f(x) = x3 + 6x2 ++8x (d) f(x) = x2 + x+ 1(e) f(x) = 2x2 − x − 1 (f) f(x) = x2 + x − 3

Exercise 8.13. What does the problem of finding the x-intercept ofa function have to do with the title, “Points of Intersection of Curves,”of this section?

� Determining the intersection of two Curves. Consider the two func-tions y = f(x) and y = g(x). We wish to find all points, if any, on theintersection of the graphs of f and g.


Let (x0, y0) be a point that is on both graphs of f and g. This means

f(x0) = y0 and g(x0) = y0.

At this point, we have f(x0) = y0 = g(x0). This represents a criterionfor finding the points of interection of two graphs.

Procedure. Let y = f(x) and y = g(x) be two function. Set up theequation

f(x) = g(x) (6)

and solve for x.

Finding the points of intersection is essentially a problem in solvingequations.

Example 8.4. Find the points of interection beween(a) f(x) = 3x+ 2 and g(x) = 5x − 4(b) f(x) = x2 − 3x+ 1 and g(x) = 2x − 5


Notice that the points of intersection can be calculated without ref-erence to the graph of the functions. At our level of play, finding thepoints of intersection is purely an exercise in algebra.

Exercise 8.14. Find the cartesian coordinates of the points of inter-sections of each of the following pairs of functions.(a) f(x) = 6x+ 3 and g(x) = 2x − 7(b) f(x) = x+ 3 and g(x) = 2 − 8x(c) f(x) = x2 + 7x − 1 and g(x) = 4x − 3

Now for some exercises that require the use of the Quadratic For-mula.

Exercise 8.15. Find the abscissas of intersection of each of the fol-lowing pairs of functions.(a) f(x) = 2x2 − 5x+ 2 and g(x) = x+ 3(b) f(x) = x2 + 4x − 1 and g(x) = 1 − 4x − x2

(c) f(x) = 3x2 + 1 and g(x) = x2 − 5x


Some curves do not intersect. Investigate these kind.

Exercise 8.16. Verify algebraically that each pair of functions donot intersect.(a) f(x) = 4x − 2 and g(x) = 4x+ 12(b) f(x) = 2x2 + 3 and g(x) = x2 − 1(c) f(x) = x2 + 2x+ 2 and g(x) = x+ 1(d) f(x) = x2 − 2x − 4 and g(x) = 4x2 − 3

We have come to the end of Lesson 8. Congratulations of reachingthis far. In Lesson 9, we take up the topics of linear and quadraticfunctions as well as some graphing topics.


8.1. Solutions: I’ll just use equation (1).(a) P (0, 0) and Q(3,−4)

d(P,Q) =√(0 − 3)2 + (0 − (−4))2 =

√9 + 16 =

√25 = 5

(b) P (−3, 4) and Q(−1,−1)

d(P,Q) =√(−3 − (−1))2 + (4 − (−1))2

=√(−2)2 + 52 =

√4 + 25

=√29

(c) P (−4, 2) and Q(5, 2)

d(P,Q) =√(−4 − 5)2 + (2 − 2)2

=√(−9)2 = | − 9| � Recall,

√x2 = |x|.

= 9


(d) P (−1, 5) and Q(7, 9)

d(P,Q) =√(−1 − 7)2 + (5 − 9)2

=√(−8)2 + (−4)2 =

√64 + 16

=√80 = 4

√5

Exercise 8.1.


8.2. Solutions:(a) Given P (0, 0), Q(1, 1) and R(2, 0), does PQR form a right tri-

angle?Solution: Verify the following calculation:

d(P,Q) =√2 d(P,R) = 2 d(Q,R) =

√2

Note that PR is the longest side and

d(P,R)2 = d(P,Q)2 + d(Q,R)2

Thus, the points P , Q, and R form a right triangle.(b) Given P (6,−7), Q(11,−3) and R(2,−2), does PQR form a right

triangle?Solution: Verify the following calculations:

d(P,Q) =√41 d(P,R) =

√41 d(Q,R) =

√82

The side QR is the longest side. Note that

d(Q,R)2 = d(P,Q)2 + d(P,R)2


Thus, the points P , Q, and R form a right triangle.(c) P (1, 2), Q(−3, 4) and R(4,−2)

Solution: Verify the following calculations:

d(P,Q) =√20 = 2

√5 d(P,R) = 5 d(Q,R) =

√85

The side QR is the longest side. Note that

d(Q,R)2 = 85 �= 45 = 20 + 25 = d(P,Q)2 + d(P,R)2

The square of the longest side does not equal to the sum of thesquares of the other two sides; therefore, P , Q and R do notform a right triangle.

Exercise 8.2.


8.3. Solutions:(a) Given P (1, 3), Q(2, 5) and R(4, 9), is Q between P and R?

Solution:

d(P,Q) =√5 d(P,R) =

√45 d(Q,R) =

√20

(Verify these calculations.) Is it true that

d(P,R) ?= d(P,Q) + d(Q,R)√45 ?=

√5 +

√20 � Not obvious! Simplify!

3√5 ?=

√5 + 2

√5

They are equal! Indeed, Q does lie between P and R.(b) Given P (−1, 10), Q(2,−5), and R(5,−12), is Q between P and

R?Solution:

d(P,Q) =√234 d(P,R) =

√520 d(Q,R) =

√58


(Verify these calculations.) Is it true that

d(P,R) ?= d(P,Q) + d(Q,R)√520 ?=

√234 +

√58

Make a calculator calculation to see that√520 �=

√234 +

√58

In this case, the left hand side is not equal to the right handside. Conclusion: Q is not between P and R.

Exercise 8.3.


8.4. Solution: The perimeter of a triangle is the sum of the lengthsof its sides.

Verify the following calculations, and e-mail if I am in error.

d(P,Q) =√17 d(Q,R) =

√5 d(R,P ) = 2

√2

The perimeter is

perimeter =√17 +

√5 + 2

√2

Question. Is this triangle a right-triangle?(a) Yes (b) No

Exercise 8.4.



8.5. Solution: I’ll take my own hint—I hope you did too.

Given: P (1 + t, 3 − t) and O(0, 0)

d(P,O) =√(1 + t)2 + (3 − t)2 � distance formula

We want d(P,O) = 4, therefore,

Solve for t:√(1 + t)2 + (3 − t)2 = 4

Square both sides of the equation, expand and combine.

(1 + t)2 + (3 − t)2 = 16 � square both sides

(1 + 2t+ t2) + (9 − 6t+ t2) = 16 � expand

2t2 − 4t+ 10 = 16 � combine

t2 − 2t+ 5 = 8 � divide both sides by 2

t2 − 2t − 3 = 0 � subtract 18 from both sides


We now solve the equation t2 − 2t − 3 = 0. We could use the Qua-dratic Formula; however, this is an equation that can be solved byfactoring:

t2 − 2t − 3 = 0

(t − 3)(t+ 1) = 0Therefore,

t = −1, 3

Presentation of Solution: t = −1, 3

There are actually two times at which the particle is exactly 4 unitsaway from the origin. Exercise 8.5.


8.6. Demonstration: Since the points are vertically oriented, x1 = x2:

d(P,Q ) =√(x1 − x2)2 + (y1 − y2)2

=√(y1 − y2)2 � since x1 = x2

= | y1 − y2 | � by (1) of Lesson 2

Exercise 8.6.


8.7. Solution: We use equation (3) throughout these solutions.(a) Given: P (1, 2) and Q(1, 9), compute d(P,Q). The first coordi-

nates are equal; therefore, these are vertically oriented points.The distance between them is the upper most minus the lowermost:

d(P,Q) = |2 − 9| = 9 − 2 = 7

(b) Given: P (−3, 3) and Q(−3,−4), compute d(P,Q). These pointsare vertically oriented because the x-coordinates are equal.

d(P,Q) = |3 − (−4)| = 7

(c) Given: P (4, 3) and Q(−3, 3), compute d(P,Q). Here, the secondcoordinates are equal; these are horizontally oriented points.

d(P,Q) = |4 − (−3)| = 7

(d) Given: P (π,−5) and Q(π,−2), compute d(P,Q). Vertically ori-ented points.

d(P,Q) = | − 5 − (−2)| = 3


What do you know about that, they’re all 7 units apart! Oops! Allbut one—how did that one get in there! Exercise 8.7.


8.8. Solutions: Hopefully you used formula (4).(a) Find the midpoint between P (−1, 3) and Q(5, 7).

M

(−1 + 52

,3 + 72

)= M(2, 5)

(b) Find the midpoint between P (2, 4) and Q(2,−5).

M

(2 + 22

,4 + (−5)

2

)= M(2,− 1

2 )

Did you note that these two points were vertically oriented? Youdid, didn’t you.

(c) Find the midpoint between P (5,−3) and Q(12,−3).

M

(5 + 12

2,−3 + (−3)

2

)= M( 17

2 ,−3)

These were two horizontally oriented points.


(d) Find the midpoint between P (−1,−1) and Q(4, 2).

M

(−1 + 42

,−1 + 2

2

)= M( 3

2 ,12 )

Exercise 8.8.


8.9. Solutions:(a) Find the center and radius of the circle containing the points

P (1, 2) and Q(−3,−1), which are diametrically opposite.Calculation of the Center : The radius is the midpoint of the linesegment PQ, since the points lie of the same diameter.

C(1 + (−3)

2,2 + (−1)

2

)= C

(−1,

12

)� midpoint formula

Radius Calculation: The radius is one-half the diameter. Thus,from the distance formula, we have

r =12d(P,Q) =

12

√(1 − (−3))2 + (2 − (−1))2

=12

√42 + 32 =

12

√25 =

12

· 5 = 52.


Center: C(−1,

12

)Radius: r =

52


(b) Find the center and radius of the circle containing the pointsP (−1, 4) and Q(0, 0), which are diametrically opposite.Calculation of the Center : Again, the center is midway betweenthe endpoints of any of its diameters.

C(−1

2,42

)= C

(−12, 2

)� midpoint formula

Calculation of the radius: The radius is one-half the diameter.By the distance formula, we have

r =12d(P,Q) =

12

√(−1)2 + 42 =

12

√17.

Presentation of Answers:

Center: C(−12, 2

)Radius: r =

12

√17

Exercise 8.9.


8.10. Solutions: Just use the replacement technique. Be sure to verifythese calculations.(a) Given f(x) = 2x2 − 3x; evaluate f(2), f(−2), f(− 1

2 ).

f(2) = 2 · 22 − 3 · 2 = 2

f(−2) = 2(−2)2 − 3(−2) = 14

f(− 12 ) = 2(− 1

2 )2 − 3(− 1

2 ) = 2

(b) Given g(s) = s(s+ 1)(s+ 2); evaluate g(0), g(1), g(−1), g(−3).

g(0) = 0

g(1) = (1 + 1)(1 + 2) = 6

g(−1) = 0

g(−3) = (−3)(−3 + 1)(−3 + 2)

= −3(−2)(−1) = −6


(c) Given h(t) =t

t2 + 1; evaluate h(1), h(−2), h( 1

2 ), h(− 12 ).

h(1) =1

1 + 1=

12

h(−2) =−2

(−2)2 + 1=

−24 + 1

= −25

h( 12 ) =

12

( 12 )

2 + 1=

12

14 + 1

=1254

=4

2 · 5 =410

=25

h(− 12 ) =

− 12

(− 12 )

2 + 1= −

12

14 + 1

= −25

(d) Given D(w) = w√w; evaluate D(9); D( 1

9 ).

D(9) = 9√9 = 9 · 3 = 27 D( 1

9 ) =19

√19 = 1

9 · 13 = 1

27

Exercise 8.10.


8.11. Solution to: (a) Find the domain of f(x) = |x − 1|. Here, thenatural domain is all of R, the real number line.

Dom(f) = R.

This is because there are no constraints on the value of x. For any xwe can calculate x − 1 and then calculate its absolute value |x − 1|.


Solution to (b) Find the domain of g(x) =3x

x2 + 2x − 8. In this prob-

lem, we just must make sure the denominator is never equal to zero.

Constraints on domain: x2 + 2x − 8 �= 0

To identify these x’s, we find where x2 + 2x − 8 = 0.

Solve: x2 + 2x − 8 = 0 � given

(x+ 4)(x − 2) = 0 � factor

x = −4, 2 � the solutions


Dom(g) = {x | x2 + 2x − 8 �= 0 } � initial description

= {x | x �= −4 and x �= 2 } � set notation

= (−∞,−4) ∪ (−4, 2) ∪ (2,+∞) � interval notation


Solution to (c) Find the domain of h(x) =1√

x(x+ 2). Based on the

strategy, we see there are a number of constraints on the values of x:

x �= 0 x �= −2 x(x+ 2) ≥ 0

These three can be summarized by a single inequality:

Constraints on domain: x(x+ 2) > 0

We use the Sign Chart Method to analyze x(x+ 2).The Sign Chart of x(x+ 2)

−2x+ 2

0x

−2 0x(x+ 2)




Dom(h) = {x | x(x+ 2) > 0 } � initial description

= {x | x < −2 or x > 0 } � set notation

= (−∞,−2) ∪ (0,+∞) � interval notation


Solution to (d) Find the domain of p(x) =√

x

x2 − 1. From the basic

strategy, we see thatx

x2 − 1≥ 0 x2 − 1 �= 0

Realizing that x2 − 1 �= 0 is equivalent to x �= −1 and x �= 1 is see . . .

Constraints on domain:x

x2 − 1≥ 0 x �= −1 x �= 1

Begin by doing a Sign Chart Analysis onx

x2 − 1.

The first step is to factor completely.x

x2 − 1=

x

(x − 1)(x+ 1)

It is to this expression that we now apply the method.


The Sign Chart ofx

(x − 1)(x+ 1)

0x

−1x+ 1

1x − 1

−1 0 1

x

(x − 1)(x+ 1)legend : • negative (−)

• positive (+)Taking the blue line as our solution, and keeping in mind that x �= −1and x �= 1 we get

Dom(p) = {x | x

x2 − 1≥ 0, x �= −1, x �= 1 } � initial description

= {x | −1 < x ≤ 0 or x > 1 } � set notation

= (−1, 0] ∪ (1,+∞) � interval notation

Exercise 8.11.


8.12. Solutions:(a) f(x) = 4x − 1

Set up: f(x) = 04x − 1 = 0 � substitute

Solve: 4x = 1 � add 1 to both sides

x =14

� divide by 4

Presentation of Answer : The x-intercept is x =14

(b) f(x) = x2 − 3x+ 2Set up: f(x) = 0

x2 − 3x+ 2 = 0 � substitute

Solve: (x − 2)(x − 1) = 1 � factor

x = 1, 2 � solved!

Presentation of Answer : The x-intercepts are x = 1, 2


(c) f(x) = x3 + 6x2 ++8xSet up: f(x) = 0

x3 + 6x2 + 8x = 0 � substitute

Solve: x(x+ 2)(x+ 4) = 1 � factor

x = 0, −2, −4 � solved!

Presentation of Answer : The x-intercepts are x = 0, −2, −4

(d) f(x) = x2 + x+ 1Set up: f(x) = 0

x2 + x+ 1 = 0 � substitute (quadratic equation)

Solve: 12 − 4(1)(1) < 0 � negative discriminant

No Solutions

Presentation of Answer : The function does not cross the x-axis.


(e) f(x) = 2x2 − x − 1Set up: f(x) = 0

2x2 − x − 1 = 0 � substitute

Solve: (−1)2 − 4(2)(−1) = 9 > 0 � positive discriminant

x =1 ± √

94

� quadratic formula

=1 ± 34

� simplify

= − 12 , 1

Note: This equation could have also been solved by factoringthe left-hand side.

Presentation of Answer : The x-intercepts are x = −12, 1


(f) f(x) = x2 + x − 3Set up: f(x) = 0

x2 + x − 3 = 0 � substitute

Solve: 12 − 4(1)(−3) = 13 > 0 � positive discriminant

x =−1 ± √

132

� quadratic formula

Presentation of Answer : The x-intercepts are x =−1 ± √

132

Exercise 8.12.


8.13. Answer : The x-intercept of a function y = f(x) is the inter-section the graph of f and the x-axis. The x-axis is the graph of thefunction g(x) = 0. Thus the x-axis is the intersection of two curves:y = f(x) and g(x) = 0. Exercise 8.13.


8.14. Solutions: Hopefully, you used standard procedures.(a) Find points of intersection: f(x) = 6x+ 3 and g(x) = 2x − 7.

Set up: f(x) = g(x) � equate ordinates

6x+ 3 = 2x − 7 � substitute

Solve: 4x = −10 � add −2x − 3 both sides

x = − 104 � divide by 4

= − 52 � done!

When x = − 52 , f(− 5

2 ) = 6(− 52 ) + 3 = −12.

Presentation of Answer : P (− 52 ,−12 )

(b) Find points of intersection: f(x) = x+ 3 and g(x) = 2 − 8x.


x+ 3 = 2 − 8x � substitute

Solve: 9x = −1 � add 8x − 3 both sides

x = − 19 � divide by 9


When x = − 19 , f(− 1

9 ) = − 19 + 3 = 26

9 .

Presentation of Answer : P (− 19 ,

269 )

(c) Find points of intersection: f(x) = x2+7x−1 and g(x) = 4x−3.


x2 + 7x − 1 = 4x − 3 � substitute

Solve: x2 + 3x+ 2 = 0 � add −4x + 3 both sides

(x+ 1)(x+ 2) = 0 � factor

x = −1, −2 � done!

Calculation of Ordinates: When x = −1, y = g(2) = −7.When x = −2, y = g(3) = −11.


Points of Intersection: P1(−1,−7 ), P2(−2,−11 )

Exercise 8.14.


8.15. Solutions: We use standard procedures around here . . . howabout you?(a) Find points of intersection: f(x) = 2x2−5x+2 and g(x) = x+3.


2x2 − 5x+ 2 = x+ 3 � substitute

Solve: 2x2 − 6x − 1 = 0 � add −x − 3 both sides

(−6)2 − 4(2)(−1) = 44 > 0 � pos. discrim.

x =6 ± √

444

� Quadratic formula

=3 ± √

112

These two curves intersect at x =3 − √

112

,3 +

√11

2


(b) Find points of intersection: f(x) = x2 + 4x − 1 and g(x) =1 − 4x − x2.


x2 + 4x − 1 = 1 − 4x − x2� substitute

Solve: 2x2 + 8x − 2 = 0 � add −1 + 4x + x2

82 − 4(2)(−2) = 80 > 0 � pos. discrim.

x =−8 ± √

804


=−8 ± 4

√5

4= −2 ±

√5

These two curves intersect at x = −2 −√5,−2 +

√5


(c) Find points of intersection: f(x) = 3x2 + 1 and g(x) = x2 − 5x.


3x2 + 1 = x2 − 5x � substitute

Solve: 2x2 + 5x+ 1 = 0 � add −1 + 4x + x2

52 − 4(2)(1) = 17 > 0 � pos. discrim.

x =−5 ± √

174


These two curves intersect at x =−5 − √

174

,−5 +

√17

4

Exercise 8.15.


8.16. Solutions: Follow the procedure.(a) f(x) = 4x − 2 and g(x) = 4x+ 12


4x − 2 = 4x+ 12 � substitute

Solve: 0 = 14 � add −4x + 2 both sides

The equation 0 = 14 has no solution; i.e., no value of x cansatisfy the equation 0 = 14. Therefore, these two curves do notintersect.

(b) f(x) = 2x2 + 3 and g(x) = x2 − 1.


2x2 + 3 = x2 − 1 � substitute

Solve: x2 = −4 � add −x2 + 1 both sides

The equation x2 = −4 has no solutions; therefore, these twocurves do not intersect.


(c) f(x) = x2 + 2x+ 2 and g(x) = x+ 1


x2 + 2x+ 2 = x+ 1 � substitute

Solve: x2 + x+ 1 = 0 � add −x − 3 both sides

(1)2 − 4(1)(1) = −3 < 0 � negative discriminant

A negative discriminant (b2−4ac < 0) implies that the equationhas not solution; therefore, these two equations do not intersect.

(d) f(x) = x2 − 2x − 4 and g(x) = 4x2 − 3


x2 − 2x − 4 = 4x2 − 3 � substitute

Solve: −3x2 − 2x − 1 = 0 � add −4x2 + 3

(−2)2 − 4(−3)(−1) = −8 < 0 � negative discriminant

A negative discriminant implies the equation has no solution.Exercise 8.16.


8.1. Solution:P : (x1, y1) = (−2, 4)

Q : (x2, y2) = (3,−1)

We take the difference in the first coordinates and the difference inthe second coordinates.

x1 − x2 = −2 − 3 = −5

y1 − y2 = 4 − (−1) = 4 + 1 = 5

We now take the sum of the squares of these two:

(x1 − x2)2 + (y1 − y2)2 = (−5)2 + 52 = 25 + 25 = 50.

Finally, we take the square root of this result:

d(P,Q) =√(x1 − x2)2 + (y1 − y2)2 =

√50 = 5

√2

Presentation of Solution: d(P,Q) = 5√2


Of course, this process can be accelerated once you fully understandthe computational steps. Example 8.1.


8.2. Solution to: (a) Define f(x) = x2+3x+1. The natural domain isthe set of all real numbers for which the value of f(x) = x2+3x+1 canbe computed as a real number. For any real number x, the expressionx2 + 3x+ 1 evaluates to a real number. Therefore, we deduce,

Dom(f) = R = (−∞,+∞)

Solution to: (b) Define g(x) =x2

x2 − 3x+ 2. The numerator and de-

nominator always evaluate to a real number; however, if the denom-inator evaluates to zero, the quotient is not a real number. Thus, wecan say that

Dom(g) = {x | x2 − 3x+ 2 �= 0 }This should not be considered to be a satisfactory characterization ofthe domain of g though.


First find where x2 − 3x+ 2 = 0, and reason from there. Solve

x2 − 3x+ 2 = 0 � given

(x − 1)(x − 2) = 0 � factor

x = 1, 2 � solve

Therefore,

Dom(g) = {x | x2 − 3x+ 2 �= 0 }= {x | x �= 1, x �= 2, } � set notation

= (−∞, 1) ∪ (1, 2) ∪ (2,+∞) � interval notation

Solution to: (c) Define h(x) =√x+ 2. For any x, x + 2 evaluates to

a real number, but for√x+ 2 to evaluate to a real number we must

have x+ 2 ≥ 0. Thus,

Dom(h) = {x | x+ 2 ≥ 0 }


Again, we should not be satisfied with this formulation. We next solvethe inequality:

x+ 2 ≥ 0 =⇒ x ≥ −2

Thus,Dom(h) = {x | x ≥ −2 } = [−2,+∞)

Solution to: (d) Define p(x) =√x2 − 1. In order for p(x) to evaluate

to a real number, we require x2 − 1 ≥ 0 and x �= 0. Thus,

Dom(p) = {x | x2 − 1 ≥ 0 }We need to solve the inequality x2 − 1 ≥ 0. To do this, we use theSign Chart Method originally discussed in Lesson 7. (Actually, thismethod really isn’t needed for this simple inequality. We could solveas follows:

x2 − 1 ≥ 0 =⇒ x2 ≥ 1 =⇒ |x| ≥ 1,

but we shall use the Sign Chatr Method in any case, just to remindyou of this method.)


We begin by factoring completely the left-hand side (which is a dif-ference of squares):

(x+ 1)(x − 1) ≥ 0

The Sign Chart of (x+ 1)(x − 1)

−1x+ 1

1x − 1

−1 1(x+ 1)(x − 1)


Therefore, the solution to the inequality x2 − 1 ≥ 0 is

(−∞,−1 ] ∪ [ 1,+∞ )

But the solution to this inequality is the natural domain of p. Thus,

Dom(p) = (−∞,−1 ] ∪ [ 1,+∞ )


Notice how all the techniques of algebra (Lessons 1–7) are used:factoring, solving inequalities, interval notation and so on.

This is the discouraging and challanging thing about mathematics:To solve any given problem, we must call on our entire history ofexperiences in mathematics. This is why it is so important for us totry to master each of the little steps we take toward our final goals.

Example 8.2.


8.3. Solution: Consider f(x) =√

x

x+ 1. Based on the above strat-

egy, we see that

Dom(f) ={x | x �= −1 and

x

x+ 1≥ 0

}

The first condition, x �= −1, avoids having zero in the denominator (weexclude x = −1 from the domain); the second condition is necessaryfor the radicand to be nonnegative. (The square root of a nonnegativenumber is a real number, whereas the square root of a negative numberis a complex number. We don’t want to work with complex numbersat this time.)

As you can see, I’ve simply translated the strategy into a series ofinequalities. We solve the inequality

x

x+ 1≥ 0

first using the Sign Chart Methods.


The Sign Chart ofx

x+ 1

−1x+ 1

0x

−1 0

x

x+ 1legend : • negative (−)

• positive (+)Therefore,

x

x+ 1≥ 0 =⇒ x ≤ −1 or x ≥ 0.

But, we also have the condition x �= −1; this changes the above solu-tion slightly to

x < −1 or x ≥ 0.


Dom(f) = (−∞,−1) ∪ [0,+∞)


Example 8.3.


8.4. Solutions: We follow the standard procedures.(a) Find the points of intersection of f(x) = 3x+2 and g(x) = 5x−4.


3x+ 2 = 5x − 4 � substitute

Solve: −2x = −6 � add −5x − 2 to both sides

x = 3 � divide by 3

At x = 3, f(3) = 3(3) + 2 = 11. Thus, ( 3, 11 ) is the point ofintersection.Presentation of Solution: Intersection Point(s): P ( 3, 11 )


(b) Find intersection points of f(x) = x2−3x+1 and g(x) = 2x−5.


x2 − 3x+ 1 = 2x − 5 � substitute

Solve: x2 − 5x+ 6 = 0 � add −2x + 5 both sides

(x − 2)(x − 3) = 0 � factor

x = 2, 3 � done!

Calculation of Ordinates: When x = 2, y = g(2) = −1.When x = 3, y = g(3) = 1.


Points of Intersection: P1( 2,−1 ), P2( 3, 1 )

Example 8.4.

Important Points

Why Abscissa and Ordinate?

This terminology enables us to refer to the horizontal and vertical axes(and the first and second coordinates) in a manner that is independentof the labeling of the axis. Sometimes the axes are labeled by usingother letters such as s, t, u or v; in these cases, when we discuss thex-axis, for example, we may not have an x-axis.

Of course, the letters x and y are symbols representing the horizontaland vertical axes. Whatever we say about the “x-axis” we are sayingabout the horizontal axis, whatever its name.

Sometimes it is convenient to refer to the axes without referring to aspecific coordinate axis label. Important Point


The correct answer is (b): Q is between P and R if and only if thedistance from P to R equals the combined distances from P to Q andQ to R. This statement is summarized in the equation

d(P,R) = d(P,Q) + d(Q,R).

Comments: Statements (a) and (c) imply the three points form aright triangle . . . that’s not what we want. Statement (d) means R isbetween P and Q . . . close but not what we wanted either.

To understand the solution, draw a picture of three colinear points inthe plane. Label the two extreme points P and R and the one betweenthem Q. Observe that d(P,R) = d(P,Q) + d(Q,R).

To understand my comments, draw pictures in the plane to reflecteach alternative. Important Point


Solutions to the Quiz. The basic tool is the horizontal distanceformula (2).

1. Yes. Since the ordinates of the two points (2, 6) and (−4, 6) areequal, this means, by (2), they are horizontally oriented.

2. Answer (c). The distance between the points P (3,−9) andQ(8,−9) is

d(P,Q) = |x1 − x2| = |3 − 8| = | − 5| = 5

3. Answer (d). The distance between the points P (6,−3) andQ(−3,−3) is

d(P,Q) = |x1 − x2| = |6 − (−3)| = |6 + 3| = 9

Notice the use of the parentheses to properly evaluate the for-mula.

4. Answer (a). The distance between the points P (−5, 2) andQ(−9, 2) is

d(Q,P ) = |x1 − x2| = | − 5 − (−9)| = | − 5 + 9| = |9 − 5| = 4


Again, note the use of parentheses to properly evaluate the for-mula. Of course, d(P,Q) = d(Q,P )—I hope that didn’t botheryou.

Did you get 100%? I hope so. Important Point


Let the Cartesian coordinates of M be denoted by M(x, y). We cancompute the values of x and y by simply describing the geometricproperties of M :

d(P,Q) = d(P,M) + d(M,Q) � M is between P and Q

d(P,M) = 12d(P,Q) � M is halfway between

We have two equations and two unknowns (x and y), we can solve forx and y. However, this method is rather messy.

The enthusiastic student can pursue this train of thought.Important Point


Models for Functions.In this section we present different ways of thinking about functionsthat may be of help to you.

� A Function as a Mapping.One traditional way of looking at a function is as a mapping or atransformation. Let f :A → B be a function, and let x ∈ A. As dis-cussed above, y = f(x) is the value of the function at x. We can alsolook upon f as a mapping or transformation: f maps x onto y, or, yis the image of x under f .

This interpretation is one of the origins of the notation introducedabove:

xf�→ y.

Try to get the feeling for this interpretation. Imagine a bunch of arrowspointing from elements x in the set A to elements y in the set B. Thearrows point from each x to the corresponding value of y, as the “arrownotation” above suggests. When we see x we immediately think of its


corresponding value f(x). The Venn Diagram, described next, is amore visual representation.

� Venn Diagram of a Function.

Figure I-1

In Figure I-1, a pictorial representation of a function (map-ping, transformation) is given. This graph represents f as itmaps or transforms a typical element x from the domain set

A into the co-domain set B. The image of x under this map, f , is de-noted by y in the figure. Visualize a function as a bunch of “arrows”pointing from set A into set B. The tail of a typical arrow is at x, andthe arrow “points” to the corresponding y-value.

This model is very useful in understanding functions and various op-erations performed on functions (such as composition of functions).

Figure I-2

To further illustrate the point, Figure I-2 depicts a relationthat is not a function. A function is a rule that associateswith each value x is a certain domain set, a corresponding

unique y-value. A rule that associates with at least one x more thanone corresponding y-value would not be a function—as illustrated


in Figure I-2. Observe that associated with x is two correspondingvalues—labeled y and z.

As a particular example of this, consider the equation: x2 + y2 = 1.For x = 0, there are two values of y that satisfy this equation: y = 1and y = −1. This equation does not define, therefore, y as a functionof x. (Visualize two arrow coming out of x = 0, one pointing to y = 1and the other pointing to y = −1.

� A Function as a Black Box.This interpretation of function is often associated with the engineeringworld. A function is like a machine (a black box). We have a machine(a black box) that takes input into it, and, as a result, yields output.The black box is the function, the input are the values in the domainof the function, and the output of the box (function) are the valuesin the range of the function.

x −→ function −→ y.

Actually, this looks more like a white box to me :={).


A black box you are familiar with is the hand-held calculator. Thisis usually, literally, a black box. You input x-values on the key pad,say x = 12. You then choose the black box to input this value of xinto. Your calculator is actually made up of a large number of blackboxes—called function keys (Hey, function!). Choose the function keylabeled

x2

and press it – out comes the output. You will see (on your real orimagined display panel) the value 144.

This is a representation of the black box model.

x −→ x2 −→ x2,put x = 12,

12 −→ x2 −→ 144.

Input-output, input-output – and that’s the way it works.Important Point

��mptiimenu


Lesson 9: Functions (cont.) & First Degree Curves

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 9: Functions (cont.) & First Degree Curves

Table of Contents1. Functions (cont.) & First Degree Curves1.1. Functions Revisited

• Is y a function of x? • Graphing • The Vertical Line Test1.2. Lines

• The Slope of a Line • The Two-Point Form • The Point-Slope Form • The Slope-Intercept Form • The General Form• Parallel & Perpendicular Lines

9. Functions (cont.) & First Degree Curves

9.1. Functions Revisited

• Is y a function of x?In everyday conversational speech we say that “something is a func-tion of something else.” For example, “success is a function of effort,”“postage on a letter is a function of its weight,” “the diameter of atree is a function of its age,” and so on. These are statements assertthat a measure of one quantity, in some way, depends on the measureof another quantity. However, in conversational speech, the exact wayin which one quantity depends on the other is not specified; indeed,the exact relationship may be unknown. These are usually statementsof “feelings” or “relationships.”

In mathematics, we have the same type statements, “y is a functionof x,” but here, the meaning is precise.

Section 9: Functions (cont.) & First Degree Curves

We say that y is a function of x if we can find a function f(x) suchthat

y = f(x).

There is no particular significance to the letters x and y; likewise, wecan say that x is a function of y provided there is some function g(y)such that

x = g(y).

What does it mean for w to be a function of s? It means that thevariable w is expressible in terms of s:

w = h(s),

for some function h(s).

Illustration 1. Here are some examples of functions of differentvariables.(a) y = 4x3 − 3x + 1 defines y as a function of x.(b) x = 5y4 − 8y + 12 defines x as a function of y.(c) w = 2t + 1 defines w as a function of t.(d) t = 1

2 (w − 1) defines t as a function of w.


(e) A = πr2 defines A, area, as a function of r, the radius of a circle.(f) C = 2πr defines circumference of a circle, C, as a function of r,

the radius.

(g) r =C

2πdefines radius, r, as a function of C.

Example 9.1. Given y = 3x − 1. Write x as a function of y.

Exercise 9.1. Respond to each of the following.(a) Given y = 5x + 1, write x as a function of y.(b) Given 4x − 2y = 1, write y as a function of x.(c) Given 4x − 2y = 1, write x as a function of y.(d) Given 2w + 3s3 = 3, write s as a function of w.(e) Given 2w + 3s3 = 3, write w as a function of s.

In applications, you must create your own function of the appropriatetype.

Exercise 9.2. Suppose we have a cube and the total surface area isknown. The problem is to find the length of the common side of the


cube. Your assignment : Write the length, x, of the side of a cube as afunction of the surface area, S; i.e., write x = f(S). Use your formulato compute the length of the common side of a cube having a totalsurface area of 24 square units.

Exercise 9.3. Write the radius, r, of a sphere as a function of itsvolume, V .

Exercise 9.4. Write the radius, r, of a sphere as a function of itssurface area, S.

Exercise 9.5. Solve each of the following: (a) A circle has a radiusof r = 3 units. Find the circumference, C, of the circle; (b) Anothercircle has a circumference of C = 20π, find the radius, r, of this circle.(Reference: Part (f) and Part (g) of Illustration 1.)

• GraphingAs you know, any equation in two variables, x and y, can be repre-sented as a curve in the plane. This curve is called the graph of theequation.


We symbolically denote an equation by

F (x, y) = c (1)

where the left-hand side, F (x, y), represents some expression in twovariables, and the right-hand side c, is some constant. The graph of(1) is the set of all points (x, y) in the plane that satisfy the equation.(Sometimes we say that the graph is the locus of all points that satisfythe given equation.)

When we identify the points on the graph of an equation, it is typi-cally a curve in the plane. The graph, then, is a visual and geometricmanifestation of the equation (1).

Figure 1

For example, consider the equation x2/3+y2/3 = 2. The point(1, 1) satisfies the equation since 12/3+12/3 = 2; in fact, all thepoints (1, 1), (−1, 1), (1,−1) and (−1,−1) satisfy the equa-

tion. These, though, are only four of infinitely many points. If weidentify enough points we can get a “feel” for the shape of the curveand are thus able to draw it in. Figure 1 represents the graph of thiscurve.


We shall not go into the technicalities of graphing; in the age of thegraphing calculator, many of the these age-old techniques seem anti-quated. Rather, we shall concentrate on study in this lesson, on theproperties of the straight line. In Lesson 10 we shall look at ad-ditional curves: the parabola, the circle, and an introduction to thecircular (or trig) functions.

Some curves are the graphs of functions, others are not. The firstorder of business is to classify curves as functional curves or not. Inthe next section, we develop a basic, yet important, criteria for doingexactly this.

• The Vertical Line TestCurves that are graphs of functions are particularly important. Lety = f(x) be a function. (In this case, we say that y is a function ofx.) We may look upon y = f(x) as an equation as discussed above.(Rewrite y = f(x) as y−f(x) = 0, this makes is look like (1); however,it is usually inconvenient to write it this way.) The fact that y is afunction of x means that for any choice of x = x0 in the domain of f ,


there corresponds a unique y = y0. This (x0, y0) pair will satisfy theequation because that is the definition of a function: If y0 correspondsto x0 with respect to the function f , then y0 = f(x0).

Now for the problem: Suppose you have a curve C drawn in the xy-plane. How can we tell whether this curve C represents y as a functionof x?

There is a simple graphical test.

Vertical Line Test :A curve C in the xy-plane defines y as a function of x if it istrue that every vertical line intersects the curve at no morethan one point.

Important. The x-axis is assumed to be the horizontal axis, and sothe meaning of vertical is perpendicular to the x-axis.


Exercise 9.6. Taking the definition of function into consideration,the orientation of the axes (x-axis is horizontal), and the geometry ofthe graph of a curve, justify in your own mind the Vertical Line Test.

The equation y = 2x + 1 defines y as a function x—here, f(x) =2x + 1. Usually, we like the independent variable to be on the x-axis(the horizontal axis); sometimes, however, things do not work out asplanned. The equation x = y5 − y defines x as a function of y.

The concept of function is independent of the letters chosen to expressthe relationship; here, in this instance, x = y2 − y does define x as afunction of y—for each value of y there corresponds only one x. Nowthe independent variable y is one the vertical axis and the dependentvariable is on the horizontal axis. I hope this does not disturb youpsychologically too much.

Exercise 9.7. When x is a function of y, x = g(y), and we graph theequation in the xy-plane, what are the distinguishing characteristicsof the curve that determine it to be the graph of a function on y?


Given a curve in the plane we can discern whether it is a graph of afunction using the Vertical Line Test and its variant, the HorizontalLine Test.

Quiz. Answer each of the following questions. Passing is 100%.

Figure 2

1. Does this curve define y a function of x?(a) Yes (b) No

Figure 3


Figure 4


Figure 5

4. Does this curve define x a function of y?(a) Yes (b) No

End Quiz.


9.2. Lines

Two points determine a line. We all know that. The problem is howto express a line mathematically in a way we can manipulate it alge-braically.

Let’s dispatch two simple cases first.

� Vertical Lines: Let � be a vertical line. Characteristic of the pointson � is that they all have the same first coordinate. Let a be the x-intercept of the line �. Then an equation that describes the line � isgiven by

x = a. (2)

Thus, � is the set of all points that satisfy the above equation; i.e., allpoints having an x-coordinate of a.

For example, the equation x = 3 is the vertical line that crosses thex-axis at 3.

� Horizontal Lines: Let � be an horizontal line. Characteristic of thepoints of � is that they all have the same second coordinate. Let b be


the y-intercept of the line �, then the equation that describes the line� is given by

y = b. (3)

The line � consists of all points that satisfy the equation; i.e., all pointswhose second coordinate is b.

For example, y = −4 is the horizontal line crossing the y-axis at −4.

For nonvertical lines, the analysis is not quite so simple. We begin bya discussion of the slope of a line.

• The Slope of a LineLet � be a nonvertical line. Choose any two distinct points on �; callthem P (x1, y1) and Q(x2, y2). Then P and Q are not vertically ori-ented.

Question. What does it mean for P and Q not to be vertically ori-ented?(a) x1 = x2 (b) x1 �= x2 (c) y1 = y2 (d) y1 �= y2


Slope of a Line:Let � be a nonvertical line and let P (x1, y1) and Q(x2, y2)be any two points on this line. The slope, m, of the line � isgiven defined to be

m =y2 − y1

x2 − x1. (4)

Comments: The value of the slope, m, does not depend on the twopoints chosen to compute it. Indeed, if P (x1, y1) and Q(x2, y2)

Figure 6are chosen from the line, and if P ′(x′

1, y′1) and Q′(x′

2, y′2) are

another pair of points from the line, then by the principle ofsimilar triangles Figure 6, we have

y2 − y1

x2 − x1=

y′2 − y′

1

x′2 − x′

1. (5)


We compute the slope by taking the difference in the ordinates of thetwo points and dividing by the difference in their abscissa—begin sureto subtract in the same order.

Example 9.2. Calcuate the slope of the line that passes throughP (−1, 3) and Q(4, 1).

Exercise 9.8. Find the slope of the line passing through each of thegiven pair of points.(a) P (3, 1) and Q(4, 5) (b) P (−2,−3) and Q(0, 2)(c) P (− 1

2 , 23 ) and Q(−3, 2) (d) P (2, 1) and the origin

• The Two-Point FormNow let’s turn to the problem of creating an equation that describesa line.

Given two points P (x1, y1) and Q(y1, y2). These two points determinea unique line, namely the line passing through P and Q; call this line


�. The slope of the line is given by

m =y2 − y1

x2 − x1(6)

As was remarked earlier, the slope calculation does not depend on theparticular points on the line �. If we take a generic point R(x, y) fromthe line it should be true that

m =y − y1

x − x1. (7)

That is, m is also determined by the points R and P .

Now equate equations (6) and (7) to obtain

y − y1

x − x1=

y2 − y1

x2 − x1or,

y − y1 =y2 − y1

x2 − x1(x − x1)

This last equation, called the two-point form of the equation of a line,characterizes the points on the line �.


Two-Point Form:Let P (x1, y1) and Q(x2, y2) be two nonvertical points in theplane. Then the equation of the line passing through thesetwo points is given by

y − y1 =y2 − y1

x2 − x1(x − x1) (8)

Example 9.3. Find the equation of the line passing through the twopoints P (−2, 4) and Q(6, 9).

Having obtained the equation of a line, we can start to extract infor-mation from it.

Example 9.4. For the line obtained in Example 9.3, find the y-intercept, the x-intercept, and the points on the graph correspondingto x = −3 and x = 6.


Exercise 9.9. Find the equation of the line passing through the twopoints P (−4,−1) and Q(5, 1).

Exercise 9.10. Given the equation developed in the Exercise 9.9,find the y-intercept and the x-intercept.

Exercise 9.11. Show that the equation for the line that crosses thex-axis at x = 2 and the y-axis at y = 5 can be written in the form

x

2+

y

5= 1.

Exercise 9.12. (The Two-Intercept Form) Suppose a line crossesthe x-axis at x = a and the y-axis at y = b. Show that the equationfor this line can be written as

x

a+

y

b= 1 (9)


The two-intercept form can be a very quick way of writing down anequation for a line if you know both intercepts.

Exercise 9.13. Write the equation of the line that crosses the x-axisat x = 4 and the y-axis at y = −3.

• The Point-Slope FormThe two-point form was introduced above as a transitional form; i.e.,you use it to establish an initial equation and then you manipulatethe equation to put it into a more useful form. The point-slope formis the same way.

The Point-Slope Form:Suppose it is known that a line � has slope m and passesthrough a point P (x1, y1). The equation of the line � is givenby

y − y1 = m(x − x1) (10)


Example 9.5. Find the equation of the line with a slope of m = 5and passes through the point P (1, 2). Leave the answer in the formof y as a function of x.

Exercise 9.14. Find the equation of the line that passes throughthe point P (3,−6) and having slope m = −2. Leave your answer inthe form of y as a function of x. Find another point on the line andusing these two points, make a sketch of the line.

Exercise 9.15. Find the equation of the line with a slope of m = − 12

that passes through the point P (−2,−9). Leave your answer in theform of x as a function of y.

• The Slope-Intercept FormThe slope-intercept form is one of the final forms of an equation of aline, as opposed to a transitory form. The point-slope form is y−y1 =


m(x − x1). As we did in the examples and exercises in the previousparagraphs, we wrote this equation in the form of a function of x:

y = y1 + m(x − x1)= y1 + mx − mx1

= mx + (y1 − mx1)= mx + b � put b = y1 − mx1

The equation y = mx + b is called the slope-intercept form of theequation of a line. Of course, m is the slope of the line. But whatis the interpretation of the number b? The y-intercept of a line isobtained by putting x = 0 and solving for y. If we do that here we get

y = mx + b |x=0 = b

Thus, y = b is the y-intercept of the line.

Let’s record this in shadow box form.


Slope-Intercept Form:The equation of the line � having a slope of m and a y-intercept of b given by

y = mx + b (11)

Exercise 9.16. Find the equation of the line that has slope 6 andcrosses the y-axis at −1.

Exercise 9.17. A line crosses the y-axis at y = 6 and has a slope ofm = −2. Find the equation of the line and leave your answer in theform of a function of x.

Exercise 9.18. Find the equation of the line that passes throughthe two points P (3, 1) and Q(4, 5). Leave your answer in the infamousslope-intercept form, y = mx + b. (Reference: two-point form.)


Exercise 9.19. A line has slope m = 12 and passes through the point

P (−2,−5), find the equation of this line and leave your answer in theslope-intercept form; i.e., leave your answer in the form of y as afunction of x. (Reference: The point-slope form.)

Quiz. Copy down the equation of the line described in Exercise 9.19and use it to answer each of the following questions. Passing is 100%.1. What is the y-intercept of this line?

(a) −8 (b) −4 (c) 4 (d) 82. Which of the following is the x-intercept of the line?

(a) −8 (b) −4 (c) 4 (d) 83. Is the point (2,−1) on this line?

(a) Yes (b) No4. Write the equation for the line with x as a function of y:

(a) x = 2y + 8 (b) x = 2y − 8 (c) x = 12y + 4 (d) x = 1

2y − 45. Use the results of Question #4 to obtain the abscissa (the

x-coordinate) of the line when the line has an ordinate (a y-coordinate) of 10.


(a) 1 (b) 10 (c) 14 (d) 28EndQuiz.

This all seems all very simple. Given certain information, it is easy tocompute the equation of the line.

Exercise 9.20. Consider the two lines y = 3x − 1 and y = 1 − x.(a) Compute the slopes of each of these two lines(b) State the y-intercepts of each line.

Exercise 9.21. The two lines in Exercise 9.20 intersect each other.Find the point of intersection.

• The General FormAll the forms for the equation of a line can ultimately be written as

Ax + By + C = 0 (12)

This is called the General Form for the equation of a line.


For example, in Exercise 9.19 we obtained the equation for a lineas y = 1

2x − 4, written in the slope-intercept form. After a few alge-braic steps, we obtain x − 2y − 8 = 0, which is in the same form asequation (12).

The General Form, like many of the forms, is not unique. For exam-ple, we can write x−2y−8 = 0 as 1

2x−y−4 = 0, or as 3x−6y−24 = 0.Usually though, we are true to our algebraic roots and remove anycommon factors (so 3x − 6y − 24 = 0 is not considered “good form”);and if the coefficients are all rational numbers, we usually clear frac-tions and write the equation with integer coefficients (so 1

2x−y−4 = 0is not a preferred form).

We can also put the constant term to the other side of the equation;thus, x − 2y = 8 would be considered in the General Form of theequation of a line.

Example 9.6. Calculate the slope of the line 2x − 3y = 5.


Exercise 9.22. Find the slope and y-intercept of each of the follow-ing equations.(a) 5x + 3y − 1 = 0 (b) −3x + 12y + 13 = 0 (c) x − y = 0

Exercise 9.23. Sketch the graph of each line in Exercise 9.22 byplotting the y-intercept and by finding and plotting an additionalpoint.

Exercise 9.24. Find the equation of the line that passes through thepoints (−4, 1) and (5, 1). Leave your answer in the General Form.

Recall, the y-intercept is obtained by putting x = 0 and solving fory, and the x-intercept is obtained by putting y = 0 and solving for x.Use these simple criteria to solve the next problem.

Exercise 9.25. Find the x- and y-intercepts of each of the lines,and sketch their graphs by plotting the intercepts and drawing a linethrough them.(a) 3x + 4y = 24 (b) 5x − 2y = 10 (c) x − 2y = 1


Exercise 9.26. Find the points of intersection between the followingpairs of lines. (Warning: In one of the parts below, the lines do notintersect.) Leave your answer should be a point in the plane: P (a, b).(a) x − y = 1 and x + y = 1 (b) x − 3y = 1 and 2x − y = 1(c) 4x − 2y = 7 and y = 2x + 1 (d) 6x + y = 3 and x + y = 2

• Parallel & Perpendicular LinesThe last topic under consideration in this lesson is to develop criteriafor determining whether two lines are parallel or perpendicular.

� Parallel Lines: Two lines �1 and �2 are parallel if and only if they donot intersect. Suppose the equations of these lines are y = m1x + b1and y = m2x + b2, respectively. When we try to find the point ofintersection we equate

m1x + b1 = m2x + b2

and try to solve for x:

(m1 − m2)x = b2 − b1.


We can divide by m1−m2 to get the solution for x, provided m1−m2 �=0. It is the singular case of m1 − m2 = 0 in which we cannot solve forx; this is the case in which the lines are parallel.

Parallel Lines:Two lines �1 and �2 having slopes m1 and m2 respectively,are parallel if and only if

m1 = m2 (13)

that is, if and only if they have the same slope.

When the lines are written as functions of x, i.e., in the slope-interceptform, it is trivial to see whether the lines are parallel. For example,y = 2x − 5 and y = 2x+12 are parallel because the slope of each lineis m = 2. The two lines y = 3x − 4 and y = 5x + 1 are not parallelbecause the first line has slope m1 = 3 and the second line has slopem2 = 5.


When the lines are in general form some slight effort is needed todetermine whether two lines are parallel—simply put the equationsin the infamous slope-intercept form, at which point you can make aneasy determination.

Quiz. Which of the following pairs of lines are parallel. Passing is100%.1. Are the lines x + 2y + 3 = 0 and 3x + 6y + 1 = 0 parallel?

(a) Yes (b) No2. Are the lines 3y − 2x = 3 and 12y − x = 6 parallel?

(a) Yes (b) No3. Are the lines 5x − 2y = 1 and 4y − 10x = 2 parallel?

(a) Yes (b) NoEndQuiz.

Exercise 9.27. Find the equation of the line that is parallel to theline y = 2x − 1 but passes through the point P (2, 7).

Exercise 9.28. Find the equation of the line that passes throughthe point P (−3,−1) and is parallel to the line 5x + 4y = 12.


� Perpendicular Lines: Two lines �1 and �2 are perpendicular ororthogonal if they intersect and at the point of intersection the anglebetween the two lines is 90◦. We state the following criteria with proof.

Perpendicular Lines:Let �1 be a line with slope m1 �= 0 and �2 be a line withslope m2 �= 0, then �1 is perpendicular to �2 (�1 ⊥ �2) if andonly if

m1 · m2 = −1 or m2 = − 1m2

, (14)

that is, if and only if the slope of one line is the negativereciprocal of the other.

Exercise 9.29. Think about the cases excluded explicitly in the cri-teria for perpendicular lines. When are lines that fall into the excep-tional case perpendicular?


Quiz. Which of the following pairs of lines are perpendicular to eachother. Passing is 100%.1. Are the lines y = 2x + 1 and y = − 1

2x − 3 perpendicular?(a) Yes (b) No

2. Are the lines x − 3y + 3 = 0 and 3x+ y − 2 = 0 perpendicular?(a) Yes (b) No

3. Are the lines 3x − 6y = 1 and 9x + 3y = 2 perpendicular?(a) Yes (b) No

EndQuiz.

Exercise 9.30. Find the equation of the line that passes throughthe point P (3, 4) and is perpendicular to the line y = 5x + 1.

Exercise 9.31. Find the equation of the line that passes throughthe point P (−5, 3) and is perpendicular to the line 6x + 2y = 1.

We have now developed enough tools to successfully tackle the follow-ing problem.


Exercise 9.32. Find the distance the point P (−1,−1) is away fromthe line y = 2x + 4. Answer this question by first mapping out astrategy, then carrying out your strategy. Parts (a) and (b) correspondto each of these two steps.(a) Plot the point and the line, then map out in a series of steps

how you plan to solve this problem.(b) Now carry out your game plan.

We have come to the end of Lesson 9. Lesson 10 continues the dis-cussion of some common, yet important curves: parabolas, circles, andtrig functions.


9.1. Solutions:(a) Given y = 5x + 1, write x as a function of y.

Answer : x = 15 (y − 1)

(b) Given 4x − 2y = 1, write y as a function of x.

Answer : y = 12 (4x − 1)

(c) Given 4x − 2y = 1, write x as a function of y.

Answer : x = 14 (1 + 2y)

(d) Given 2w + 3s3 = 3, write s as a function of w.

Answer : s = 3

√3 − 2w

3


(e) Given 2w + 3s3 = 3, write w as a function of s.

Answer : w = 32 (1 − s3)

Exercise 9.1.


9.2. Solution: As the first step, find surface area, S, as a function ofx. Why? Because its easy!

S = 6x2

But we don’t want this. We want x as a function of S. No problem:

x =

√S

6

Now, given that we have a cube with surface area of S = 24, we cannow easily compute the length of the common side:

x =

√S

6

∣∣∣∣∣S=24

=

√246

=√4 = 2

(Here, the vertical bar, |, is the so-called, evalutional notation—a stan-dard way of conveying the idea that we are substituting a particularvalue into a formula.) Thus,

S = 24 =⇒ x = 2

Exercise 9.2.


9.3. Solution: It is well know that V = 43πr3. This represents V as

a function of r. We want r as a function of V . Thus,

r = 3

√3V4π

Given the volume, V , this formula lends itself to the easy calculationof the radius. Exercise 9.3.


9.4. Solution: It is well known that S = 4πr2. This represents S asa function of r. We want r as a function of S. Thus,

r =

√S

4π

Given the surface area, S, this formula lends itself to the easy calcu-lation of the radius. Exercise 9.4.


9.5. Solution to (a) From Part (f), we have C = 2πr. This defines Cas a function of r; what this means is that this equation is designedto give back certain information, C, given that you know the value ofr—and we do.

C = 2πr|r=3 = 2π(3) = 6π

A circle of radius r = 3 has a circumference of C = 6π.

Solution to (b) From Part (g), we have r =C

2π. This defines r as a

function of C.

Now we are given that C = 20π; therefore,

r =C

2π

∣∣∣∣C=20π

=20π2π

= 10

The radius of a circle with circumference C = 20π is r = 10Exercise 9.5.


9.6. I cannot justify it in your own mind, only in mine.Exercise 9.6.


9.7. For each y we can have at most one x; therefore, if we draw ahorizontal line at position y on the y-axis, the line should intersectthe graph of x = g(y) in at most one location.

This is the horizontal analog to the Vertical Line Test. We can for-mulate this into a test procedure.

Horizontal Line Test :A curve C in the xy-plane defines x as a function of y if itis true that every horizontal line intersects the curve at nomore than one point.

Exercise 9.7.


9.8. Solutions:(a) P (3, 1) and Q(4, 5)

m =5 − 14 − 3

= 4

(b) P (−2,−3) and Q(0, 2)

m =2 − (−3)0 − (−2)

=52

(c) P (− 12 , 2

3 ) and Q(−3, 2)

m =2 − 2

3

−3 − (− 12 )

=43

− 52

= − 815

(d) P (2, 1) and the origin, i.e., take Q(0, 0).

m =1 − 02 − 0

=12

Exercise 9.8.


9.9. Solution: Substituting into equation (8), to obtain

Given: P (−4,−1) and Q(5, 1)

y − y1 =y2 − y1

x2 − x1(x − x1)

y − (−1) =1 − (−1)5 − (−4)

(x − (−4))

y + 1 =29(x + 4)

or,

y =29(x + 4) − 1

Exercise 9.9.


9.10. Solution: The y-intercept is found by putting x = 0 and solvingfor y:

y =29(x + 4) − 1

∣∣∣∣x=0

=29(4) − 1 = −1

9

The y-intercept is y = −19

The x-intercept is found by putting y = 0 and solving for x:

29(x + 4) − 1 = y = 0

29(x + 4) = 1

x + 4 =92

x =92

− 4 =12

The x-intercept is x =12


Thus, this line crosses the x-axis at x = −1/9 and crosses the y-axisat y = 1/2. It is now a trivial matter to draw the graph of the line.

Exercise 9.10.


9.11. Solution: The x-intercept is x = 2 and the y-intercept is y =5; this means the line passes through the points P (2, 0) and Q(0, ).Therefore,

Given: P (2, 0) and Q(0, 5)

y − y1 =y2 − y1

x2 − x1(x − x1)

y = −52(x − 2) � substitute in

Now we try to put the last equation in the prescribed form.

y = −52(x − 2) � starting point

5x + 2y = 10 � clear fractions and expand

5x10

+2y10

= 1 � divide by 10

x

2+

y

5= 1 � the required equation!

Exercise 9.11.


9.12. Solution: The solution follows along the same lines as theprevious exercise.

Given: P (a, 0) and Q(0, b)

y − y1 =y2 − y1

x2 − x1(x − x1) � two-point form

y − 0 =b − 00 − a

(x − a) � substitute in

y = − b

a(x − a) � simplify

Now we try to put the last equation in the prescribed form.bx + ay = ab � clear fractions and expand

bx

ab+

ay

ab= 1 � divide by ab

x

a+

y

b= 1 � the required equation!

Exercise 9.12.


9.13. Solution: We use the two-intercept form:

x

4+

y

−3= 1

x

4− y

3= 1

That was quick! Exercise 9.13.


9.14. Solution: We use equation (10):

Given: m = −2 and P (3,−6)

y − y1 = m(x − x1) � equation (10)

y − (−6) = −2(x − 3) � substitute

y + 6 = −2(x − 3) � a rewrite

y = −6 − 2(x − 3) � write as a function of x

y = −2x � combine

Presentation of Answer : y = −2x

The choice of a second point is entirely up to you. I’ll take x = 1.

x = 1 =⇒ y = −2x|x=1 = −2

Thus the point Q(1,−2) lies on the equation. Now plot the given pointP (3,−6) and the computed point Q(1,−2), and draw a straight linethrough these two points. Exercise 9.14.



Given: m = −12and P (−2,−9)

y − y1 = m(x − x1) � equation (10)

y − (−9) = −12(x − (−2)) � substitute

y + 9 = −12(x + 2) � a rewrite

y = −9 − 12(x + 2) � write as a function of x

y = −12x − 10 � combine

Thus, y = − 12x − 10 is the equation of the target line written as a

function of x. It was requested that the answer be written in the formof x as a function of y.

y = −12x − 10 =⇒ x = −2(y + 10)


Presentation of Answer : x = −2(y + 10) Exercise 9.15.


9.16. Solution: We are given the slope, m = 6, and the y-intercept;this is a job for the slope-intercept form of the equation of a line.

Given: m = 6 and b = −1y = mx + b � slope-intercept form

y = 6x − 1 � substitute and done!

Presentation of Answer : y = 6x − 1

Comments: Be neat and organized in your presentation. It is impor-tant to acquire a mathematical literacy. It is more difficult for me totype out a neat, well-organized presentation than it is for you to writeone out. DPS

Exercise 9.16.


9.17. Solution: This is a job for the slope-intercept form of the equa-tion of a line.

Given: m = −2 and b = 6y = mx + b � slope-intercept form

y = −2x + 6 � substitute and done!

Presentation of Answer : y = −2x + 6 Exercise 9.17.


9.18. Solution: We are given two points. We could use the two-pointform, or we could (1) compute the slope, then (2) use the point-slopeform. I’ll take the latter tack:

1. Compute the slope.

Given: P (3, 1) and Q(4, 5),

m =5 − 14 − 3

= 4

2. Now use the point-slope form.

Given/Known: P (3, 1) and m = 4,

y − y1 = m(x − x1) � point-slope form

y − 1 = 4(x − 3) � substitute

y = 1 + 4x − 12 � expand

y = 4x − 11 � combine

Presentation of Answer : y = 4x − 11 Exercise 9.18.


9.19. Solution: Given P (−2,−5) and m = 12 ,


y − (−5) = 12 (x − (−2)) � substitute

y + 5 = 12 (x + 2)

y = −5 + 12x + 1 � expand


Presentation of Answer : y = 12x − 4 Exercise 9.19.


9.20. Solution:(a) Compute the slopes of each of these two lines.

• The slope of the line y = 3x − 1 is m = 3, the coefficientof the x term.

• The slope of the line y = 1 − x is m = −1, the coefficientof the x term.

(b) State the y-intercepts of each line.• The y-intercept of the line y = 3x − 1 is b = −1, theconstant term.

• The y-intercept of the line y = 1−x is b = 1, the constantterm.

Exercise 9.20.


9.21. Solution: The point of intersection (x, y) satisfies both equa-tions. Generally, to find the point of intersection we equate the y-values:

3x − 1 = y = 1 − x � equate ordinates

3x − 1 = 1 − x � eliminate redundant notation

4x = 2

x =12

The two lines intersect when the abscissa is x = 12 . But

x =12

=⇒ y = 1 − x|x= 12= 1 − 1

2=

12

Thus, the two lines intersect at P ( 12 , 1

2 )

Draw these two lines on the same sheet of paper in order to see thepoint of intersection. (Or, have your student held graphing calculatordo it for you.) Exercise 9.21.


9.22. Solution: We simply put each in the slope-intercept form.(a) Find the slope and y-intercept of the line 5x + 3y − 1 = 0.

5x + 3y − 1 = 03y = −5x + 1

y = − 53x + 1

3 � y = mx + b

This line has slope m = − 53 and y-intercept of y = 1

3 .(b) Find the slope and y-intercept of the line −3x + 12y + 13 = 0.

−3x + 12y + 13 = 012y = 3x − 13

y = 14 − 13

12 � y = mx + b

This line has slope m = 14 and has y-intercept of y = − 13

12 .


(c) Find the slope and y-intercept of the line x − y = 0.

x − y = 0y = x � y = mx + b

This line has slope m = 1 and y-intercept of y = 0.Exercise 9.22.


9.23. Solution: Trivial. The key point is finding an additional pointon the each line—that’s done using the equation. Exercise 9.23.


9.24. Solution: There are two solutions.

The Quick Solution: Note that the two given points, (−4, 1) and (5, 1),have the same ordinate; therefore, this is a horizontal line. Its equationis

y = 1

The Long Solution: The long solution uses standard methods.

Calculate Slope: m =y2 − y1

x2 − x1

m =1 − 15 + 4

= 0

Point-Slope Form: y − y1 = m(x − x1)

y − 1 = 0(x − (−4))y = 1

Again we arrive at the conclusion that y = 1 is the equation of theline. Which way did you do it? Exercise 9.24.


9.25. Solution: The calculation of the intercepts from the generalform is very quick. The graphing is trivial so I’ll not bother to preparethe graphs for you.(a) Find the intercepts of 3x + 4y = 24.

3x + 4y = 24y-intercept: Put x = 0

4y = 24y = 6

x-intercept: Put y = 03x = 24x = 8

Presentation of Answer : The x- and y- intercepts are, respec-tively, x = 8, y = 6 .

The solutions to the other two parts are on the next page. Given thatyou have looked at the solution to part (a), you may want to reviseyour solutions to parts (b) and (c) before looking.


(b) Find the intercepts of 5x − 2y = 10.

5x − 2y = 10y-intercept: Put x = 0

−2y = 10y = −5

x-intercept: Put y = 05x = 10x = 2

Presentation of Answer : The x- and y- intercepts are, respec-tively, x = 2, y = −5 .

(c) Find the intercepts of x − 2y = 1.

x − 2y = 1y-intercept: Put x = 0

−2y = 1

y = −12

x-intercept: Put y = 0x = 1


Presentation of Answer : The x- and y- intercepts are, respec-tively, x = 1, y = − 1

2 .

The graphing of these equations is left to the student—that’s you.Exercise 9.25.


9.26. Solutions:(a) Find the point of intersection: x − y = 1 and x + y = 1.

Solution: Begin by solving for y in each equation.x − y = 1 � given

y = x − 1 � solve for y

x + y = 1 � given

y = 1 − x � solve for y

Now equate ordinatesx − 1 = y = 1 − x

and solve for xx − 1 = 1 − x � from above

2x = 2x = 1 � solved!

Now, substitute the value of x = 1 into any of the two equationsto get y = 0.

Presentation of Answer : Point of intersection is P (1, 0)


(b) Find the point of intersection: x − 3y = 1 and 2x − y = 1.

x − 3y = 1 � given

y = 13 (x − 1) � solve for y

2x − y = 1 � given

y = 2x − 1 � solve for y

Now equate ordinates

13 (x − 1) = y = 2x − 1

and solve for x

13 (x − 1) = 2x − 1 � from above

x − 1 = 6x − 3 � clear fractions

x = 25 � solve for x

Now, substitute the value of x = 25 into any of the two equations

to get y = − 15 .

Presentation of Answer : Point of intersection is P ( 25 ,− 1

5 )


(c) Find the point of intersection: 4x − 2y = 7 and y = 2x + 1.

4x − 2y = 7 � given

y = 12 (4x − 7) � solve for y

y = 2x + 1 � given


12 (4x − 7) = 2x + 1

and solve for x

12 (4x − 7) = 2x + 1 � from above

4x − 7 = 4x + 2 � clear fractions

−7 = 2 � substract 4x from both sides

Conclusions: The last equation has no solution—the x’s havebeen totally eliminated. This means that there are no x’s thatsatisfy the equation 1

2 (4x − 7) = 2x + 1; hence, these two linesdo not intersect—they are in fact parallel.


(d) Find the point of intersection: 6x + y = 3 and x + y = 2.

6x + y = 3 � given

y = −6x + 3 � solve for y

x + y = 2 � given

y = 2 − x � solve for y


−6x + 31 = y = 2 − x

and solve for x

−6x + 3 = 2 − x � from above

−5x = −1

x = 15

Now, substitute the value of x = 15 into any of the two equations

to get y = 95 .

Presentation of Answer : Point of intersection is P ( 15 , 9

5 )

Exercise 9.26.


9.27. Solution: This problem is no different from the problems seenearlier (See, e.g., Exercise 9.14); the only difference is that we getour slope from the given line.

The slope of the line y = 2x−1 is m = 2. We want to construct a lineparallel to this line, so it too should have a slope of m = 2; however,our line is required to pass through the point P (2, 7):

Given: m = 2 and P (2, 7)



y = 2x + 3

Presentation of Answer : The line that is parallel to the line y = 2x−1and passes through the point P (2, 7) is

y = 2x + 3

Exercise 9.27.


9.28. Solution: Same problem as just solved, the only difference iswe have to work a little harder to get the slope.

Given line: 5x + 4y = 124y = −5x + 12

y = − 54x + 12

The slope of the given line is m = − 54 .

Given: m = − 54 and P (−3,−1)


y + 1 = − 54 (x + 3) � substitute

y = − 54x − 15

4 − 1

y = − 54x − 19

4

Presentation of Answer : y = − 54x − 19

4 or 5x + 4y + 19 = 0Exercise 9.28.


9.29. When m1 = 0 (m2 = 0), then line �1 (resp. �2) is a horizontalline. In this case, only vertical lines are perpendicular to it!

Typically, horizontal lies have equation y = a, and vertical lines haveequation x = b. It is very easy to determine whether any line is per-pendicular to either a horizontal line or a vertical line.

Exercise 9.29.


9.30. Solution: The slope of the given line is mgiven = 5 and so theslope of the line we are trying to construct has slope

m = −15

� from (14)

We now go into our point-slope mode (our most powerful mode ofthought): Given m = − 1

5 and P (3, 4),


y − 4 = − 15 (x − 3) � substitute

y = 4 − 15x + 3

5

y = − 15x + 23

5 � slope-intercept

Presentation of Solution: The desired line is y = − 15x + 23

5

Exercise 9.30.


9.31. Solution: The given line 6x + 2y = 1 can be rewritten in theslope-intercept form as y = −3x + 1

6 . Therefore, the given line hasslope of mgiven = −3, so the line we are trying to construct has slope

m = − 1−3

=13

� from (14)

We use the point-slope form: Given m = 13 and P (−5, 3),


y − 3 = 13 (x + 5) � substitute

y = 3 + 13x + 5

3

y = 13x + 14

3 � slope-intercept

Presentation of Solution: The desired line is y = 13x + 14

3

Exercise 9.31.


9.32. Solution to (a) Let �1 denote the given line y = 2x + 4. Hereis an obvious sequence of steps.

1. Compute the equation of the line �2 that passes through thegiven point, P (−1,−1) and is perpendicular to the given line,�1 : y = 2x + 4. Draw a picture of this line.

2. Find the point of intersection of the two lines �1, the given line,and �2, the line perpendicular to the �1, as developed in Step 1.

3. Having found the point of intersection between �1 and �2, callthis point Q, the desired distance is d(P, Q), the distance be-tween point P and Q.

Carry out this game plan before looking at the solution to part (b)on the next page.


Step 1: Find the equation of the line perpendicular to the given line�1 and passing through the given point. The equation of that line is

y + 1 = − 12 (x + 1)

or,y = − 1

2x − 32

Verify these calculations if you missed this step.

Step 2: Find the point of intersection between �1 : y = 2x + 4 and�2 : y = − 1

2x − 32 .

2x + 4 = y = − 12x − 3

2therefore,

52x = − 11

2

x = − 115

Verify all details. Substitute this value of x into either of the twoequation to obtain

y = 2x + 4|x=−11/5 = − 25


Thus, the point of intersection between �1 and �2 is

Q(− 115 ,− 2

5 )

Step 3: Find the distance between P and Q.

Given P (−1,−1 ) and Q(− 115 ,− 2

5 ),

d(P, Q) =√(x2 − x1)2 + (y2 − y1)2

=√(− 11

5 + 1)2 + (− 25 + 1)2

=√(− 6

5 )2 + ( 3

5 )2

=√

3625 + 9

25

= 35

√5


Presentation of Answer : The distance between the point P (−1,−1)and the line y = 2x + 4 is

d(P, �1) =35

√5

Comments: That was a multistep problem! Each step, however, wasa problem type examined earlier. In mathematics, we try to outlinein our minds a plan of attack, then try to carry out the plan—beingready and willing to modify our plan or our thinking at all times.

This problem can be carried out in the abstract to obtain a nice littleformula.

For those who want to do more. Problem: Find a general formulafor the problem of finding the distance from a given point P (x0, y0)and a given line y = mx + b.

Solution: Just carry out the game plan as outlined in part (a).Exercise 9.32.


9.1. Solution:y = 3x − 1 � given

3x − 1 = y � transpose

3x = y + 1

x = 13 (y + 1) � solve for x

Thus, we have written x as a function of y:

x = 13 (y + 1)

Example 9.1.


9.2. Solution: To calculate slope, you take the difference in the or-dinates (the second coordinates) and divide by the difference in theabscissas (the first coordinates) being sure to subtract in the sameorder in both numerator and denominator.

Given: P (−1, 3) Q(4, 1)

m =3 − 1

−1 − 4= −2

5

The slope of the line is m = −25

Example 9.2.


9.3. Solution: We can substitute directly into equation (8). I shallconsider (x1, y1) = (−2, 4) and (x2, y2) = (6, 9).

Thus, from (8) we have

y − y1 =y2 − y1

x2 − x1(x − x1)

y − 4 =9 − 4

6 − (−2)(x − (−2))

y − 4 =58(x + 2)

An equation for this line is

y − 4 =58(x + 2).

If we write this equation so that y is a function of x we get

y = 4 +58(x + 2).

Example 9.3.


9.4. Solution: The equation of the line obtained in Example 9.3 is

y = 4 +58(x + 2). (S-1)

Find the y-intercept : The y-intercept is the point on the y-axis atwhich the line crosses the y-axis. Characteristic of points on the y-axis is that the x-coordinates are equal to zero. Therefore, to answerthe question we seek the point on the line for which the x-coordinateis zero.

Set x = 0 in (S-1) to obtain the corresponding value of y:

y = 4 +58(x + 2)

∣∣∣∣x=0

= 4 +58(2) =

214

The y-intercept is y =214


Find the x-intercept : The x-intercept is the point where the linecrosses the x-axis. Characteristic of points on the x-axis is that y = 0.To find the x-intercept we set y = 0 in equation (S-1) and solve for x.

4 +58(x + 2) = y � Now put y = 0, and . . .

4 +58(x + 2) = 0 � solve for x

58(x + 2) = −4

x + 2 = −325

x = −425

Thus the x-intercept is x = −425


Find the point on the curve corresponding to x = −2:

y = 4 +58(x + 2)

∣∣∣∣x=−3

= 4 +58(−3 + 2) = 4 − 5

8=

278

Thus, the point on the curve corresponding to x = −2 is (−2, 278 )

Find the point on the curve corresponding to x = 6:

y = 4 +58(x + 2)

∣∣∣∣x=6

= 4 +58(6 + 2) = 4 + 5 = 9

Thus, the point on the curve corresponding to x = 6 is ( 6, 9 )

Example 9.4.



Given: m = 5 and P (1, 2)

y − y1 = m(x − x1) � equation (10)


y = 2 + 5(x − 1) � write as a function of x


Presentation of Solution: y = 5x − 3

Comments: As you can see the point-slope was indeed a transitoryform. We used it to obtain an initial formula, then manipulated fromthere. Example 9.5.


9.6. Solution: We simply put the equation in the slope-interceptform.

2x − 3y = 5−3y = −2x + 5

y = 23x − 5

3 (S-3)

We can now see that the given equation has slope of m = 23 and, by

the way, crosses the x-axis at x = − 53 Example 9.6.

Important Points

Is x a function of y?

The answer is Yes. When we say that x is a function y we mean

x = f(y)

for some function f ; that is, x is expressible in terms of y. This meansthat corresponding to each y-value (the independent variable), thereis a unique x-value (the dependent variable). In Figure 5, for anygiven y, draw a horizontal line at y. Note that this line intersects thecurve in at most one location. This means that for any given y therecorresponds a single x-value. This is the definition of function.

Important Point


Throughout this discussion, we refer to Figure 6. The triangle PRQis similar to the triangle P ′R′Q′. Therefore the ratio of correspondingsides are equal. Thus,

d(R, Q)d(P, R)

=d(R′, Q′)d(P ′, R′)

or,y2 − y1

x2 − x1=

y′2 − y′

1

x′2 − x′

1

The last set of calculation follow form our discussion of the distancesbetween vertically and horizontally oriented points as presented inLesson 8.

Note: The above argument is not completely general. Do you see the“weakness” of the argument? Important Point


The point (2,−1) is not on the line y = 12x − 4 because it does not

satisfy the equation:12x − 4

∣∣x=2 = −2 �= −1.

Important Point

��mptiimenu


Lesson 10: Some Second Degree & Trig Curves

Directory


An

Algebra Review


a3a4 = a7 (ab)10 = a10b10


2x2 − 3x − 2 = (2x + 1)(x − 2)12x + 13 = 0 =⇒ x = −26G = { (x, y) | y = f(x) }


inTen Lessons



Lesson 10: Some Second Degree & Trig Curves

Table of Contents10. Some Second Degree & Trig Curves10.1. Parabolas

• x is of second degree • y is of second degree • IntersectingCurves

10.2. Circles10.3. The Trig Functions

• The Definitions and Consequences • Some Common Values• Graphs of the Trig Functions • The Other Trig Functions• Radian Measure versus Degree Measure

10. Some Second Degree & Trig CurvesIn Lesson 9 we gave an extensive discussion of first degree curves;these are equations having a straight line as its graph. In this lesson,we begin by exploring selected second degree curves followed by a briefsurvey remarks on trigonometric functions.

The general form of a second degree equation is

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (1)

where either A = 0 or B = 0. In our lesson, B = 01, so, in fact wewill look at equations of the form

Ax2 + Cy2 + Dx + Ey + F = 0 (2)

The graph of such an equation may be a parabola, a hyperbola, acircle, or an ellipse (or a degenerate graph). We discuss only parabolasand circles leaving the others to a (high school) course in Analytic

1B �= 0 induces a rotation of the graph.

Section 10: Some Second Degree & Trig Curves

Geometry, a course in Precalculus, or a course in Calculus andAnalytic Geometry.

Associated with any second degree equation (equation (2)) is a verystrong geometry. For example, a parabola has a focus and the parabolahas a certain reflection property that is exploited in the manufactureof flash lights, satellite dishes, parabolic mirrors, etc. These geometricproperties will not be covered in these lessons; again they are a topicof study in a course on Analytic Geometry. Instead, we will con-centrate on the mechanics: recognition, classification, location, andgraphing.

10.1. Parabolas

A parabola is described by a second degree equation (in the variablesx and y) in which one variable has power two and the other variablehas power one. (In terms of equation (2), this means A = 0 or C = 0,but not both.)

It is important to be able to recognize second degree equations whosegraph is a parabola. Study the following examples to acquire an “eye.”


Illustration 1. Parabolas.(a) The x variable has degree 2 and the y variable has degree 1. In

this case y is expressible as a function of x.1. Equational Form:

x2 − x + y = 1 3x2 − 2x + 3y = 5 y − x + 5x2 = 0

2. Explicit Form: y written (explicitly) as a function of x:

y = x2 y = 1 − x2 + x y = 13 (5x

2 + 2x)

(b) The y variable has degree 2 and the x variable has degree 1. Inthis case x is expressible as a function of y.

1. Equational Form:

x + y + y2 = 5 3x − 6y2 + y − 8 = 0 y2 = 2x + 3y

2. Explicit Form: x written (explicitly) as a function of y.

x = y2 x = 2y2 − 5y +12

x = 4y2 + 7y − 1


• x is of second degreeAn equation in which the x variable has the second degree and they variable has the first degree is a parabola that either opens up oropens down.

Figure 1(a) y = x2 Figure 1(b) y = 4− x2

Parabola Opening Up Parabola Opening Down

Notice that these two parabolas pass the vertical line test; this meansthat each defines y as a function of x. (The equations are y = x2 andy = 4 − x2.) A parabola has a vertex that represents the absoluteminimum, see Figure 1(a), or the absolute maximum, Figure 1(b),of the graph.


The skills associated with working with parabolas are as follows:• Recognition: Recognize the equation as a parabola of this

type.• Classification: Classify the parabola as one that opens up or

one that opens down.• Location: The parabola is located by calculating its so-calledvertex.

• Graphing: If needed, we need to graph the parabola.

� Recognition. We’ve already covered the case for recognition. Theimportant point is that the variable x as degree two and the variabley has degree one.

We shall satisfy ourselves by just presenting a quiz. Passing is 100%so don’t err!

• Quiz. Click on the green bullet to jump to the quiz on parabolas.


� Classification. The general form of a parabola that can be writtenas a function of x is

y = ax2 + bx + c, a = 0 (3)

We state the following without proof, though the validity of the state-ments will be made apparent later.

Classification Rule: Consider the parabola

y = ax2 + bx + c, a = 0. (4)

The parabola opens up if a > 0 and opens down if a < 0;i.e., if the coefficient of the second degree term is positive,the parabola opens up, and opens down if the coefficient ofthe second degree term is negative.

Naturally, you must strive to put your equation in the form of (4)before you can make the correct determination.


This is such a simple principle, let’s just quiz you on it.

• Quiz. Click on the green bullet to jump to the quiz on parabolas.

� Location. A parabola that is a function of x has the form

y = ax2 + bx + c, a = 0. (5)

Such a parabola opens up (if a > 0) or down (if a < 0). Anotherimportant feature of a parabola is the presence of a vertex. The vertexof the parabola in equation (5) can be determined by completing thesquare of the right-hand side.

Generally, when you complete the square of (5), you obtain an equa-tion of the form

y − k = a(x − h)2, (6)

where h and k are constants. It turns out that the vertex is locatedat coordinates V (h, k ).

Example 10.1. Find the vertex of the parabola y = x2 − 2x + 4.


The previous example was surely of Skill Level 0; the completionof the square was no problem. In the next example, the completionprocess is slightly trickier.

Example 10.2. Find the vertex of the parabola 4x2 + 3x + 2y = 3.

Before asking you to do a few, let’s summarize.

Finding the Vertex :Given a second degree equation that has been put into theform y = ax2 + bx + c, a = 0, the vertex can be obtainedby completing the square. This having been done, put theequation into standard form:

y − k = a(x − h)2 (7)

The vertex is located at V (h, k ).


Now, let’s go to the exercises. The solutions are on separate pages.You can look at them in order without seeing the solution to the nextone. Use this to refine you solution methods.

Exercise 10.1. (Skill Level 0) Use the technique of completing thesquare to put each of the parabolas into standard form. Having donethat, write the coordinates of the vertex in the form V (h, k ) and statewhether the parabola opens up or down.(a) y = x2 − 6x (b) y = 1 − 4x − x2 (c) x2 + 2x + y + 1 = 0

In the next exercise the problem of completing the square is a littletrickier, but I expect you will get a score of 100%.

Exercise 10.2. (Skill Level 0.5) Use the technique of completing thesquare to put each of the parabolas into standard form. Having donethat, write the coordinates of the vertex in the form V (h, k ) and statewhether the parabola opens up or down.(a) y = 4x2 + 2x + 1 (b) 2x2 − 3x − 2y + 1 = 0(c) x − y = 4x2 (d) 3x2 + 2x + 4y − 2 = 0


We now turn to the topic of graphing.

� Graphing.

Graphing a parabola is very simple once you have obtained some basicinformation about the parabola:

• Orientation. Does the parabola open up/down, or right/left.This information can be easily observed using the classificationrules.

• Location. The vertex determines the location of the parabola.See the discussion on finding the vertex.

Other points of interest in graphing a parabola are• Symmetry. A parabola that is vertically oriented—it opens up or

down—is symmetrical with respect to the vertical line passingthrough the vertex. This makes it even easier to graph.

• Points to Plot. It is not necessary to plot a large number ofpoints to make a rough sketch. Usually, plotting the vertex, andtwo other points to determine the “breath” of the parabola are


sufficient. If the parabola crosses the x-axis, then fixing thesepoint(s) is useful too.

•V (1, −1)

•V (1, −1)

Figure 2(a) y + 1 = (x − 1)2 Figure 2(b) y + 1 = 2(x − 1)2

These two parabolas have virtually the same equation; the only differ-ence is in Figure 2(b), the (x − 1)2 is multiplied by 2. This “scalingfactor” tends to narrow the parabola. The get a feel for the breadth ofthe parabola, simply plot two points symmetrically placed on eitherside of the vertex. The pass a parabolic curve through these threepoints. Three points and you’re done!


Exercise 10.3. Find the x-intercepts for the two parabolas in Fig-ure 2: (a) y + 1 = (x − 1)2 (b) y + 1 = 2(x − 1)2.

Exercise 10.4. Make a rough sketch of each of the following parabo-las on the same sheet of paper.(a) y = x2 + 1 (b) y = 2x2 + 1 (c) y = 1

2x2 + 1

Exercise 10.5. Make rough sketches of the graphs of each of the fol-lowing by putting each into standard form; classifying each as openingup or down; and finding the vertex of each.(a) 4x + y − x2 = 1 (b) y = −x2 − 6x + 1 (c) 2x − 3y = x2

• y is of second degreeWhen you have a second degree equation, like equation (2), and y hasdegree two with x only degree one (A = 0 and C = 0), you have aparabola that opens either to the left or to the right. Here are a fewvisuals.

x = y2 x + 2y = y2 2x + 3y + 4y2 = 1.


As in the previous case, to analyze these equations the basic skills wemust master the basic skills: recognition, classifcation, location,and graphing.

Figure 3(a) x = y2 Figure 3(b) x = 4− y2

Parabola Opening Right Parabola Opening Left

Notice the vertex is the point that is the right-most or left-most pointon the graph, depending on the orientation of the parabola.


Because of the similarity of these case with the previous (extensivelydiscussed) case, only exercises will be presented. In each exercise, ob-serve that the parabolas are of the advertised type and that theyequation define x as a function of y.

The general form of a parabola that opens to the left or right, writtenas a function of y, is

x = ay2 + by + c

Here is all the relevant information.


Analyzing Horizontally Oriented Parabolas:Consider the parabola

x = ay2 + by + c, a = 0. (8)

The parabola opens right if a > 0 and opens left if a < 0;This equation can be put into the standard form

x − h = a(y − k)2 (9)

using the method of completing the square. The location ofthe vertex is V (h, k ).

Exercise 10.6. Analyze the following parabolas by putting eachequation into standard form, equation (9), observing its orienta-tion, finding its vertex, and, finally, making a rough sketch of theparabola. (The solutions to each are given on separate pages so youcan, for example, look at the solution to (a) to refine your techniquesto solve (b)–(d).)


(a) x + 2y = y2 (b) 2x − 4y + y2 = 3(c) 2y2 − 4y + x = 0 (d) 2x + 3y + 4y2 = 1

• Intersecting CurvesQuiz. Let y = mx + b and y = ax2 + bx + c be a line and a parabola,respectively. What is the largest number of points of intersection pos-sible between these two curves?(a) 0 (b) 1 (c) 2 (d) 3

Let’s have a few “reminder” exercises in the form of solving equations.Try to graphically visualize the questions; the graphing fanatic mighteven graph all these equations . . . perhaps using a graphing calculator.

Exercise 10.7. Find the coordinates of intersection between the twogiven curves. (Recall : The Quadratic Formula.)(a) Write out the general strategy of solving these problems.(b) y = 2x + 1; y = x2 + 2x − 1 (c) y = 2x + 1; y = x2 + 4x − 1(d) y = x2 + 2; y = 2x2 − 3x + 3 (e) y = x2 + 2; y = 4 − (x − 1)2


10.2. Circles

b

C(h, k)

r

Let r > 0 be a number greater than zero and C(h, k )a point in the plane. The circle with center atC(h, k ) and radius r is the set of all points P (x, y )that are a distance of r units from the point C. Interms of symbols, a point P is on this circle if (andonly if) it satisfies the equation

d(P, C ) = r, (10)

where d(P, C ) is the distance between the points P and C.

Now, invoke the distance formula with P (x, y ) and C(h, k ) to obtain√(x − h)2 + (y − k)2 = r (11)

Square both sides to obtain the center-radius form for the equationof a circle

(x − h)2 + (y − k)2 = r2

Summary:


Center-Radius Form of a CircleThe equation of the circle having radius r > 0 and center atC(h, k ) is given by

(x − h)2 + (y − k)2 = r2 (12)

Illustration 2. Here are a few simple examples of circles already inthe center-radius form; in this case, basic information can be extractedabout the circle.(a) x2 + y2 = 1 is a circle of radius r = 1 with center at ( 0, 0 ).(b) (x − 1)2 + (y − 3)2 = 16 is a circle with radius r = 4 (r2 = 16)

and center at C( 1, 3 ).(c) (x + 2)2 + (y − 4)2 = 5 is a circle with radius r =

√5 (r2 = 5)

and center at C(−2, 4).

To construct the equation of a circle, therefore, you need two piecesof information: (1) the center of the circle and (2) the radius of the


circle. All your efforts must be concentrated towards acquiring thesetwo.

Exercise 10.8. Find the equation of the circle with center at C andpasses through the given point P(a) Write out a strategy for solving this type of problem.(b) C( 1, 2 ); P ( 4,−1 ) (c) C(−2, 3 ); P ( 5, 2 )

Exercise 10.9. Find the equation of the circle that passes throughthe two diametrically opposite points P1 and P1.(a) Write out a strategy for solving this type of problem.(b) P1( 1, 3 ) and P2( 3, 5 ) (c) P1(−3, 1 ) and P2( 4, 6 )

Normally, the circle is initially not in the center-radius form, but isoften in the general form. If you expand (multiply out) equation (12)and write it in the form of a second degree equation, equation (2), youget

Ax2 + Ay2 + Bx + Cy + D = 0, A = 0 (13)

The equation is called the General Form for the equation of a circle.


Important Points. Notice the coefficients of the two squared terms areequal (to A). This (almost) characterizes a second degree equation asbeing a circle. It is also important to emphasize that the coefficientsof the squared terms have the same sign! For example, 2x2 + 2y2 +x + y − 1 = 0 is a circle but 2x2 − 2y2 + x + y − 1 = 0 is not !

Here are a few quick visual manifestations of equation (13).

Illustration 3. Examples of Circles in General Form.(a) x2 + y2 + 2x − 3y + 1 = 0.(b) 2x2 + 2y2 − 4x + 8y + 3 = 0(c) −3x2 − 3y2 + 3x − 4y + 2 = 0. Here the common coefficient of

the squared terms is A = −3, that’s o.k. Needless to say, thisequation can be rewritten in a more esthetically pleasing form:3x2 + 3y2 − 3x + 4y − 2 = 0.

� Transforming to the Center-Radius Form. The center-radiusform is an “information yielding form” for a circle; therefore it isbeneficial to put an equation representing a circle into this form.


The Method of Transforming. Given an equation on the form of equa-tion (13), you can complete the square of the x-terms and completethe square of the y-terms to obtain the center-radius form of theequation of a circle.

Example 10.3. Put each of the following circles in the center-radiusform and find their center and radius.(a) x2 + y2 − 2x + 4y + 1 = 0 (b) 2x2 + 2y2 + 4x − y − 2 = 0

You see from the example the skills of needed to put an equation inthe center-radius form are skills you have been practicing in the pastseveral lessons. The other important skill is interpretation of results!It doesn’t do you much good if you can’t extract the information(center and radius).

I should mention, before we go to the exercises that not every equationof the form (13) is a circle. Here are three very simple examples to


illustrate the possibilities:

(1) x2 + y2 = 1 It’s graph is a circle

(2) x2 + y2 = 0 It’s graph is a point

(3) x2 + y2 = −1 This equation has no graph!

In (2), the only point (x, y) that satisfies the equation is (0, 0); hence,its graph consists of a single point. (A degenerate circle?) In (3), thereis no point (x, y) that satisfies the equation, so there are no points onthe graph.

Exercise 10.10. Put each of the following into the center-radiusform. If an equation represents a circle, state its center and radiusand if it does not represent a circle characterize its graph.(a) x2 + y2 − 6x + 2y − 4 = 0 (b) x2 + y2 + 4y = 2(c) 3x2 + 3y2 − 8x = 0 (d) x2 + y2 + 2y − 4x + 10 = 0


Exercise 10.11. Find the points of intersection between the twogiven curves.(a) x2 + y2 = 4; y = 2x (b) x2 + y2 = 4; y = x + 1(c) x2 + y2 − 2x = 4; y = 2x − 1 (d) x2 + y2 − 2x = 4; x + 2y = 6.

10.3. The Trig Functions

The trigonometric functions play an important role in many scientificand technical fields as well as in many of the trades (carpentry comesto mind). In this section we give a less than complete presentation ofsome basic ideas.

• The Definitions and ConsequencesThe definitions of the trig functions are quite simple. For any numbert, we want to define cos(t) and sin(t). This is done using the unitcircle, x2 + y2 = 1.

The Wrapping Definition of Sine and Cosine: Let t be any given realnumber. Draw the unit circle x2 + y2 = 1. Beginning at the pointI( 1, 0 ), called the initial point, measure off a length of t units around


the unit circle; if t ≥ 0, mark off this length counter clockwise andif t < 0, mark off |t| units in a clockwise direction. The point on thecircle at the end of your measurement process is called the terminalpoint, (temporarily) denoted by Pt(x, y).

bPt(x, y) t ≥ 0

bPt(x, y)

t < 0

Figure 4

If the terminal point is given by Pt(x, y), then define

cos(t) = x and sin(t) = y (14)

That is, the cos(t) is defined to be the first coordinate of the terminalpoint, and the sin(t) is defined as the second coordinate.


Let’s list off some simple consequences of this wrapping definition.

� The Fundamental Identity for Sine and Cosine:

cos2(t) + sin2(t) = 1 (15)

This is because, by our definition, the cos(t) and sin(t) are the firstand second coordinates, respectively, of the terminal point Pt(x, y).But Pt is a point on the unit circle and as such, its coordinates satisfythe equation x2 + y2 = 1; therefore, cos2(t) + sin2(t) = 1 for anynumber t.

� Symmetry Properties of Sine and Cosine:

cos(−t) = cos(t) sin(−t) = − sin(t) (16)

We say a function f(t) is an even function if it satisfies the equationf(−t) = f(t) for all t in its domain; consequently, we may say thatcos(t) is an even function. A function f(t) is called an odd function if itsatisfies the equation f(−t) = −f(t) for all t in its domain; therefore,we are entitled to say that sin(t) is an odd function.


The proof of equation (16) is given using the wrapping definition ofsine and cosine.

b

Pt(x, y)

b

P−t(x, −y)

Think of t as positive. Measure t units counterclockwise starting at I(0, 1) and terminating atPt(x, y). (This is shown in blue). By definition,cos(t) = x and sin(t) = y. Now take the nega-tive number −t and measure off t units clockwisearound the circle, shown in red, to the terminalpoint P−t(x,−y). Because we are wrapping thesame length, but in opposite directions, the x-

coordinate of P−t will be the same as the x-coordinate of Pt; they-coordinate of P−t will be of negative the y-coordinate of Pt. Sincethe terminal point of −t is P−t(x,−y), by definition we have thatcos(−t) = x and sin(−t) = −y; but x = cos(t) and y = sin(t). Puttingthese observations together we get

cos(−t) = x = cos(x) sin(−t) = −y = − sin(t),

which are the symmetry properties.


� Periodicity of Sine and Cosine:

cos(t + 2π) = cos(t) sin(t + 2π) = sin(t) (17)

In this case we say that cos(t) and sin(t) are periodic of period 2π. (Ingeneral, a function f(t) is periodic of period p provided f(t+p) = f(t)for all t in its domain. Not all functions are periodic.)

Again, this property is best seen through a picture.

Pt(x,y)

For any given number t, assumed to be posi-tive for the purposes of demonstration, wrap tunits around the unit circle, as shown in blue.Denote the terminal point by Pt(cos(t), sin(t)).Now compute the cosine and sine of the num-ber t + 2π. Begin by measuring off t + 2π unitsaround the circle; we do this by first measuringoff t units (shown in blue) followed by an addi-

tional 2π units (shown in red). Recalling that the circumference of theunit circle is 2π, we see that when we wrap the additional 2π around,


we are just going in circles! :-) Therefore, Pt is the same as Pt+2π; inwords, the terminal point for the number t is the same as the terminalpoint for the number t + 2π:

Pt(x, y) = Pt+2π(x, y).

It follows, from the wrapping definition, that

cos(t) = x = cos(t + 2π) sin(t) = y = sin(t + 2π),

which are the periodic equations.

� The Range of Cosine and Sine: We first observe that all coordinateson the unit circle are between −1 and 1. Because the cos(t) and sin(t)are coordinates on the unit circle, it follows

−1 ≤ cos(t) ≤ 1 − 1 ≤ sin(t) ≤ 1 (18)

• Some Common ValuesIn the age of the hand-held graphing calculator, why, you may ask,should I know some of the common values of the sine and cosine? Theshort answer is that some instructors, including yours truly, expect the


student to know these values. These values are among the “minimalset of knowledge” anyone talking to you about the trig functions wouldexpect you to know. You want to be at least minimal, don’t you? :-)

The circumference of the unit circle is 2π. The numbers t = 0, t = π/2,t = π, t = 3π/2, and t = 2π, when “wrapped” around the unit circle,go zero of the way around, a quarter of the way around, a half of theway around, three-quarters the way around, and all the way around,respectively. The terminal points of these five values of t can be easilyobserved:

P0(1, 0) Pπ/2(0, 1) Pπ(−1, 0) P3π/2(0,−1) P2π(1, 0) (19)

The coordinates of these “easy” points contain the cosines and sinesof the corresponding values of t. (Given in the subscript of the point,I might add.)

• Quiz. As a test of your understanding of the wrapping definition,and the meaning of the above notation, take the following quiz onsome of the common values of sine and cosine by clicking on the greenbullet.


Now let’s concentrate on three other values of t: t = π/6, t = π/4,and t = π/3. These can be worked out rather easily, though it willnot be done in these lessons:

Pπ/6(√

3/2, 1/2) Pπ/4(√

2/2,√

2/2) Pπ/3(1/2,√

3/2)

Look at the corresponding unit circle with these values wrapped.

b (√

32 , 1

2 )

b (√

22 ,

√2

2 )b ( 12 ,

√3

2 )

Figure 5(a) t = π6 Figure 5(b) t = π

4 Figure 5(c) t = π3

In Figure 5(a), we see that the cos(π/6) =√

3/2 and sin(π/6) = 1/2.These values correspond to the base and height of the right triangle.(You can see that the height of the triangle is (about) 1/2 the radiusof the unit circle.) The companion values, shown in Figure 5(c), arecos(π/3) = 1/2 and sin(π/3) =

√3/2. (Again notice the base is of the


triangle is about 1/2 the radius.) Finally, Figure 5(c) tells us thatcos(π/4) =

√2/2 and sin(π/4) =

√2/2.

Here is a table and figure summarizing the values of sine and cosine.

t cos(t) sin(t)

0 1 0

π/6√

3/2 1/2

π/4√

2/2√

2/2

π/3 1/2√

3/2

π/2 0 1b

b

b

b

b

P0(1, 0)

Pπ/6(√

32 , 1

2 )

Pπ/4(√

22 ,

√2

2 )

Pπ/3(12 ,

√3

2 )Pπ/2(0, 1)

Table 1 Figure 6

By utilizing the symmetry property, the periodic property, and thewrapping definition of sines and cosines you can easily calculate anyinteger multiple of π

n , where n = 1, 2, 3, 4, 6. In particular, it is im-portant to keep a visualization of the wrapping pictures of Figure 4.


In the example that follows, I exhibit the simple reasoning necessaryto calculate any value integer multiple of

π

n, for n = 1, 2, 3, 4, 6.

Example 10.4. Calculate the exact value of each of the following:(a) sin(−π/3) (b) cos(2π/3) (c) cos(5π/4) (d) sin(−π/2)

Tip. How to Determine Values of Sine and Cosine: To calculate thesine or cosine of multiples of π/n (n = 1, 2, 3, 4, 6) (1) draw a unitcircle; (2) plot the terminal point (this may require division, e.g.,

13π6

=136

π = 2 16 π = 2π +

π

6,

thus, 13π/6 is π/6 “beyond” 2π); (3) observe the relation between thispoint and one of the terminal points given in Table 1; (4) calculatevalue with the proper sign affixed.

Note. It is actually only necessary to wrap clockwise; i.e., you needonly deal with positive values of t. This is because of the symmetry


properties; thus,

sin(−7π/3) = − sin(7π/3) need only find sin(7π/3)

cos(−9π/4) = cos(9π/4) need only find cos(9π/4)

Exercise 10.12. Find the values of each of the following without theaid of a calculator.(a) sin(3π/4) (b) cos(7π/6) (c) sin(−7π/6) (d) cos(−21π/2)(e) sin(9π/3) (f) cos(10π/3) (g) sin(10π/3) (h) cos(−29π/6)

• Graphs of the Trig FunctionsHaving defined, for any number t, the cos(t) and sin(t), we now wantto study them as functions. Usually, the independent variable of afunction is x, not t. Replace the letter t by the letter x to obtain twotrigonometric functions:

y = cos(x) y = sin(x)

In this brief section, we remind the reader of the graphs of these twofunctions.


� The Graph of y = cos(x). Let us begin graphing the cosine andmaking some comments.

1 2 3 4 5 6 7 8−1−2−3−4−5−6−7−8

1

−1

Figure 7 y = cos(x)

The fact that the graph is symmetric with respect to the y-axis is amanifestation of the even function property of the cosine. The regular,repeated behavior of the graph is the periodicity of the cosine. Thecosine function repeats its pattern of values over consecutive intervalsof length 2π, the period.

The x-intercepts of the cosine are all values of x for which cos(x) = 0.Based on the wrapping definition of the cosine, it is easy to see that


the x-intercepts are

x = . . . , −5π2

, −3π2

, −π

2,

π

2,

3π2

,5π2

, . . . (20)

Or, in other words, the graph of y = cos(x) crosses the x-axis at allodd multiples of π/2. This is an important point.

The cos(x) attains its maximum value of y = 1 at

x = . . . ,−4π,−2π, 0, 2π, 4π, . . . (21)

In other words, cos(x) takes on its maximum value at all even multiplesof π.

Exercise 10.13. List, in a style similar to equation (21), the valuesof x at which the cos(x) takes on its minimum value of y = −1.Having done that, write a good English sentence that describes thesenumbers. (See the sentence that follows equation (21) above.)


� The Graph of y = sin(x). Let us begin graphing the sine and makingsome comments.

1 2 3 4 5 6 7 8−1−2−3−4−5−6−7−8

1

−1

Figure 8 y = sin(x)

The sine function is an odd function; the graph of an odd function issymmetrical with respect to the origin. The regular repeated behavioris the periodicity of the sine function. The sine function repeats itspattern of values over consecutive intervals of length 2π, the period.

Exercise 10.14. The x-intercepts. List out the x-intercepts of y =sin(x) using the style of equation (21), then describe this set of pointsusing a good English sentence. (References: Table 1 and Figure 8.)


Exercise 10.15. Maximum Values of sin(x). List the values of xat which the function y = sin(x) attains its maximum value of y = 1.Then describe these numbers in a good English sentence.

Exercise 10.16. Minimum Values of sin(x). List the values of xat which the function y = sin(x) attains its minimum value of y = −1.Then describe these numbers in a good English sentence.

� The Two Graphed Together : It may be useful to view the graphs ofthe sine and cosine on one sheet of electronic paper in order to makea direct comparison between the two.

1 2 3 4 5 6 7 8−1−2−3−4−5−6−7−8

1

−1y = cos(x)y = sin(x) Figure 9


Notice that the graphs of y = cos(x) and y = sin(x) appear to beidentical; identical in the sense that one graph, say y = cos(x), canbe obtained by shifting the graph of y = sin(x) to the left. Thisobservation is in fact correct; more precisely. The cosine graph can beobtained from the sine graph by shifting the sine graph over

π

2units

to the left.

• The Other Trig FunctionsThere are four other trig functions that hitherto I have failed to men-tion. Their definitions are in terms of cos(x) and sin(x). These fourare the tangent, cotangent, secant, and cosecant functions:

tan(x) =sin(x)cos(x)

sec(x) =1

cos(x)

cot(x) =cos(x)sin(x)

csc(x) =1

sin(x)

(22)

Of course, we make these definitions for all values of x for which thedenominators are nonzero.


The trigonometric functions play diverse roles in mathematics.• They are quite commonly used to develop relationships between

the sides of right triangles. These relationships are important inapplications in such different fields as carpentry, architecture,surveying, navigation, and computer graphics . . . to name justa few.

• In Calculus, these functions are studied like any other function.The trigonometric functions enable us to solve a greater vari-ety of problems that require the special techniques of calculus.Having trigonometric functions enables scientist and engineersto properly formulate and solve complex problems, often withthe aid of calculus and geometry.

The interesting and very useful applications must be reserved for aproper course in algebra, trigonometry, calculus, and/or courses inspecialized fields. These lessons are meant as a minimal review ofbygone memories.

Example 10.5. Identify the domain of definition of the function y =tan(x).


Quick Quiz. Is the domain of sec(x) the same as that of tan(x)?(a) Yes (b) No

Exercise 10.17. Identify the domain of definition of the functiony = cot(x). (Hint : The domain would be all values of x for which thedenominator is nonzero.)

• Radian Measure versus Degree MeasureThe wrapping definition of sine and cosine tells us how to computethe sine and cosine of a number t. The number t is not interpreted asan angle, but was simply any real number. Within the context of thewrapping definition, the number t is said to be measured in radians.

What is angle measurement? As you wrap the number t around theunit circle, you are revolving around the center of the circle The dis-tance around the circle is its circumference 2π; one complete revolu-tion around the circle is referred to as a 360◦ revolution.

In this case we say 2π radians is the same as 360◦ (degrees). Thedegree measurement of an arbitrary number t is calculated by direct


proportions; let the symbol θ denote the degree measurement of t,then using direct proportions:

θ

t=

360◦

2πthus,

θ = t · 360◦

2π= t · 180◦

π

The number 180/π is referred to as the scaling factor for convertingradians to degrees.

Converting Radians to Degrees:Let t be a number (measured in radians). The degree mea-surement of t is given by

θ = t · 180π

(23)


Illustration 4. Calculate the degree measurement of each of the fol-lowing. by applying the conversion formula, equation (23).

(a) For t =π

4, θ =

π

4180◦

π=

180◦

4= 45◦ .

(b) For t = −π

6, θ = −π

6180◦

π= −180◦

6= −30◦ .

This is not hard so I’ll just ask you to make a few calculations yourself.

Exercise 10.18. Calculate the degree measurement of each of thefollowing. by applying the conversion formula, equation (23).

(a)2π3

(b) −5π6

(c) 3π (d) −3π4

(e)5π2

In the same way we can convert degrees to radians by simply solvingthe equation (23) for t:

t = θ · π

180=⇒ t = θ · π

180.

Elevating this to the status of shadow box we obtain the following.


Converting Radians to Degrees:Let θ be a number measured in degrees. The radian mea-surement of θ is given by

t = θ · π

180(24)

Applying this formula is the same as applying the conversion for-mula (23) so no examples will be presented.

Exercise 10.19. Convert degrees to radians in each of the following.(a) 30◦ (b) 60◦ (c) 45◦ (d) −300◦ (e) 225◦

Naturally, if θ is a number measured in degree, we naturally definecos(θ) = cos(t) and sin(θ) = sin(t), where t is the radian measurementof θ, as computed by equation (24). For example,

cos(30◦) = cos(π/6) =√

32

.


Let’s revisit Table 1 and include degree measurement. Below is atable of common values of sine and cosine that you should and mustknow.

t θ cos(t) sin(t)

0 0◦ 1 0

π/6 30◦ √3/2 1/2

π/4 45◦ √2/2

√2/2

π/3 60◦ 1/2√

3/2

π/2 90◦ 0 1

Table 2

The angle measurement is usually seen in the context of problemsinvolving triangles. These problem types will not be covered in theselessons—sorry.

As a final set of thoughts, let’s connect up the geometric relationshipbetween radian measure and degree measure.


b

θ

Pt(cos(t), sin(t))

b

θ

Pt(cos(t), sin(t))

Figure 10

The degree measure is a measure of how many times we wrap aroundthe unit circle based on 360◦ rather than 2π, as is the case of radianmeasure. The student needs to be able to move quickly and quietlybetween these two systems of measurement.

This point represents the end of our Algebra Review in TenLessons. It ended not with a bang but a measure!

If you have waded through all ten lessons, my hearty congratulationsto you. There was much of algebra and trigonometry not covered in


these lessons. These electronic tomes were not meant to be an entirecourse, but a review of basic concepts and techniques.

In additional to the basic techniques and ideas presented, I hope youcome away from these lessons with the ability to use good and standardnotation and to present your thoughts in a clear and concise manner.This would be most appreciated by your professors, and is a sign ofyour mathematical maturity.

I hope these imperfect lessons have helped you review some of thebasics concepts your professors will assume you know. Now, on toe-Calculus!

To exit, either go to the top of this file and follow the arrows, or simplyclick here. DPS


10.1. Solution: We simple complete the square!

Solution to (a). Analyze: y = x2 − 6x.

y = x2 − 6x given

y = (x2 − 6x + 9) − 9 take 12 coeff. of x

y = (x − 3)2 − 9 the square is completed!

Now we put this last equation in standard form:

y + 9 = (x − 3)2

We see this is a parabola that opens down and has its vertex atV ( 3,−9 ).


Solution to (b) Analyze: y = 1 − 4x − x2.

y = 1 − 4x − x2 given. Now assoc. x terms

y = 1 + (−4x − x2) Now, factor our coeff. of x2

y = 1 − (x2 + 4x) Next, take 12 coeff. of x

y = 1 − (x2 + 4x + 4) + 4

{Adding 4 inside parenthesessame as adding 4 outside

y = 5 − (x + 2)2 basically done!

Now write the last equation in standard form:

y − 5 = −(x + 2)2

This is a parabola that opens down and has its vertex at V (−2, 5 ).


Solution to (c) Analyze: x2 + 2x + y + 1 = 0.

Begin by writing the equation as a function of x;

y = −1 − x2 − 2x

and not continue as before:

y = −1 − x2 − 2x given

y = −1 − (x2 + 2x) factor our coeff. of x2

y = −1 − (x2 + 2x + 1) + 1

{Adding 1 inside parenthesessame as adding 1 outside

y = −(x + 1)2 done!

Summary : This is a parabola that opens down having vertex locatedat V (−1, 0 ). (Here, relative to the notation of the standard form,k = 0.) Exercise 10.1.


10.2. Solution to (a) Analyze: y = 4x2 + 2x + 1.

y = 4x2 + 2x + 1 given

y = 4(x2 + 12x) + 1 factor our coeff. of x2

y = 4(x2 + 12x + 1

16 ) + 1 − 14 complete square

y = 4(x + 14 )

2 + 34 write as perfect square

Summary : The equation in standard form is

y − 34 = 4(x + 1

4 )2

This is a parabola that opens up and has vertex at V (− 14 , 3

4 ).


Solution to (b) Analyze: 2x2 − 3x − 2y + 1 = 0. Begin by solving fory as a function of x:

y = x2 − 32x + 1

2

and proceed using standard techniques.

y = (x2 − 32x) + 1

2

y = (x2 − 32x + 9

16 ) + 12 − 9

16 complete square

y = (x − 34 )

2 − 116 write as perfect square

Summary : The equation written standard form is

y + 116 = (x − 3

4 )2.

This equation describes a parabola that opens up and has its vertexat V ( 3

4 ,− 116 ).


Solution to (c) Analyze: x − y = 4x2. Begin by solving for y as afunction of x:

y = −4x2 + x

and complete the square.

y = −4(x2 − 14x) factor out leading coeff.

y = −4(x2 − 14x + 1

64 ) + 116 complete square

y = −4(x − 18 )

2 + 116 write as perfect square

This last equation written is standard form is

y − 116 = −4(x − 1

8 )2.

This is a parabola that opens down with vertex as V ( 18 , 1

16 ).


Solution to (d) Analyze: 3x2 + 2x + 4y − 2 = 0. Write as a functionof x:

y = − 34x2 − 1

2x + 12

and complete the square.

y = − 34x2 − 1

2x + 12 given

y = − 34 (x

2 + 23x) + 1

2 factor out leading coeff.

y = − 34 (x

2 + 23x + 1

9 ) + 12 + 3

4 · 19 complete square

y = − 34 (x + 1

3 )2 + 7

12 write as perfect square

Summary : Write in standard form:

y − 712 = − 3

4 (x + 13 )

2

This equation represents a parabola that opens down and has vertexat V (− 1

3 , 712 ).

Remarks: As you can see, these are all solved exactly the same way.The only difference is the level of difficulty in completing the square.


This is why fundamental algebra is so important: It is used to extractinformation. Unless you can perform the algebra, you cannot extractthe information—not good in the information age! Exercise 10.2.


10.3. Solutions(a) Find the x-intercepts of y + 1 = (x − 1)2. We use standard

methods. To find the x-intercept, we put y = 0 and solve for x.

1 = (x − 1)2 or (x − 1)2 = 1

We now solve for x:

|x − 1| = 1 take square root of both sides

x − 1 = ±1 solve absolute equation

x = 1 ± 1 solve for x

x = 0, 2 enumerate solutions

Presentation of Answer : The x-intercepts are x = 0, 2.


(b) Find the x-intercepts of y + 1 = 2(x − 1)2. Put y = 0 and solvefor x.

1 = 2(x − 1)2 or (x − 1)2 = 12

Now solve for x:

|x − 1| = 1√2

take square root of both sides

x − 1 = ± 1√2

get rid of absolute values

x = 1 ± 1√2

solve for x

Presentation of Answer : The x-intercepts are at

x = 1 − 1√2, 1 +

1√2.

Notice they are symmetrically placed on either side of the vertex.Exercise 10.3.


10.4. Solution: Your rough graphs should roughly look like the fol-lowing:

V (0, 1)

Legend : y = x2 + 1; y = 2x2 + 1; y = 12x2 + 1. Exercise 10.4.


10.5. Solutions: Here are outlines of the solutions.

(a) Analyze: 4x + y − x2 = 1.

4x + y − x2 = 1 given

y = x2 − 4x + 1 re-write

y = (x − 4x + 4) + 1 − 4 complete square

y + 3 = (x − 2)2 write in proper form

Summary : This is a parabola that opens up with vertex at V ( 2,−3 ),the sketch is left to the reader.


(b) Analyze: y = −x2 − 6x + 1.

y = −x2 − 6x + 1 given

y = −(x2 + 6x) + 1 factor out leading coeff.

y = −(x2 + 6x + 9) + 1 + 9 complete square

y − 10 = −(x + 3)2 write in proper form

Summary : This is a parabola that opens down and having vertex atV (−3, 10 ).


(c) Analyze: 2x − 3y = x2.

2x − 3y = x2 given

−3y = x2 − 2x isolate

−3y = (x2 − 2x + 1) − 1 complete square

−3y = (x − 1)2 − 1 write as perfect square

y = − 13 (x − 1)2 + 1

3 divide through

y − 13 = − 1

3 (x − 1)2 put in standard form

Summary : This is the equation of a parabola that opens down withvertex at V (1, 1

3 ).Exercise Notes: In this last part, I did something a little different.I completed the square of the l.h.s. before dividing through by thecoefficient of y; this was to avoid working with fractional expressionsunnecessarily. Many of the previous examples and exercises could havebeen handled the same way.

Exercise 10.5.


10.6. Solution to (a) Analyze: x + 2y = y2.

First write the equation as a function of y:

x = y2 − 2y.

Now, complete the square

x = (y2 − 2y + 1) − 1 take 12 coeff. of y

x = (y − 1)2 − 1 write as perfect square

Summary : The equation written in standard form is

x + 1 = (y − 1)2

This is a horizontally oriented parabola (since y has degree 2 and xhas degree 1), it opens to the right (since the coefficient of the y2 termis positive), and its vertex is located at V ( ,−1, 1 ).

The sketch of the graph is left to the student—that’s you. DPS


Solution to (b) Analyze: 2x − 4y + y2 = 3.

Begin by solving for x; writing x as a function of y:

x = − 12y2 + 2y + 3

2

Now, complete the square!

x = − 12 (y

2 − 4y) + 32 factor out leading coeff.

x = − 12 (y

2 − 4y + 4) + 2 + 32 take 1

2 coeff. of y and square

x = − 12 (y − 2)2 + 7

2


x − 72 = − 1

2 (y − 2)2.

This describes a horizontally oriented parabola that opens left withvertex at V ( 7

2 , 2 ).

The graph is left to the student, but don’t plot too many points. Threeare enough!


Solution to (c) Analyze: 2y2 − 4y + x = 0.

Write x as a function of y:

x = −2y2 + 4y

and complete the square

x = −2(y2 − 2y) factor out leading coeff.

x = −2(y2 − 2y + 1) + 2 complete the square

x = −2(y − 1)2 + 2 write as perfect square

Summary : The parabola written in standard form is

x − 2 = −2(y − 1)2.

It opens left with vertex at V ( 2, 1 ).

The graph is left to the student.

Ever get the feeling you have done this before?


Solution to (d) Analyze: 2x + 3y + 4y2 = 1.

First write the equation as a function:

x = −2y2 − 32y + 1

2

and now, complete the square,

x = −2(y2 + 34y) + 1

2 factor out leading coeff.

x = −2(y2 + 34y + 9

64 ) + 932 + 1

2 complete square

x = −2(y + 38 )

2 + 2532


x − 2532 = −2(y + 3

8 )2.

It opens to the left and has vertex at V ( 2532 ,− 3

8 ).

The graph is left to the student. Exercise 10.6.


10.7. Solution to (a) To find where two curves intersect we set the y-coordinates equal and solve for x. In this case, the resulting equationhas the second degree, so we can anticipate solving either using themethod of factoring, or, most likely, using the Quadratic Formula.


Solution to (a) Find where y = 2x + 1 and y = x2 + 2x − 1 intersect.First not that this is the intersection of a line and a parabola. Theywill either not intersect, intersect at one point, or intersect at twopoints. (Can you visualize each situation?) Let’s see which one it is inthis case.

Begin by setting the ordinates equal.

x2 + 2x − 1 = y = 2x + 1 set ordinates equal

x2 + 2x − 1 = 2x + 1 eliminate the middle man

x2 = 2 add 1− 2x to both sides

|x| =√

2 take square root both sides

x = ±√

2 solve absolute equality

Summary : The two curves intersect at two points: When the abscissasare x = −√

2 and x =√

2. “Plugging” these values of x back into thestraight line, we can calculate the coordinates of intersections:

Points of Intersection: (−√

2, 1 − 2√

2 ), (√

2, 1 + 2√

2 )


Solution to (b) y = 2x + 1; y = x2 + 4x − 1

x2 + 4x − 1 = y = 2x + 1 equate ordinates

x2 + 4x − 1 = 2x + 1 eliminate the middle man

x2 + 2x − 2 = 0 sub. 2x+ 1 from both sides

x =−2 ± √

22 − 4(−2)2

quadratic formula

=−2 ± √

122

=−2 ± 2

√3

2= −1 ±

√3

Summary : The abscissas of intersection are x = −1 ±√

3 .

The interested reader can go on to calculate the coordinates of inter-section just as I did in part (b).


Solution to (c) y = x2 + 2; y = 2x2 − 3x + 3. This is the intersectionof two parabolas that open up.

2x2 − 3x + 3 = y = x2 + 2 equate ordinates

2x2 − 3x + 3 = x2 + 2 eliminate the middle man

x2 − 3x + 1 = 0 combine similar terms on l.h.s.

x =3 ± √

32 − 42

quadratic formula

=3 ± √

52

Summary : The abscissas of intersection are3 ± √

52

.


Solution to (d) y = x2 + 2; y = 4 − (x − 1)2. These again are twoparabolas, one opening up, the other down.

x2 + 2 = y = 4 − (x − 1)2 equate ordinates

x2 + 2 = 4 − (x − 1)2 get rid of y

x2 + 2 = 3 + 2x − x2 expand r.h.s.

2x2 − 2x − 1 = 0 combine similar terms on l.h.s.

x =2 ± √

22 − 4(2)(−1)2(2)

quadratic formula

=2 ± √

124

=1 ± √

32

simplify!

Summary : The abscissas of intersection are x =1 ± √

32

.

Exercise 10.7.


10.8. Solution to (a) We are given the center C of the circle and onepoint, P , on the circle. All we need is the radius of the circle. Fromequation (10) we have

d(P, C ) = ror,

r = d(P, C ).

This boxed formula represents the method of solving these problems.


Solution to (b) We already have the center of the circle, we need theradius.

We follow the plan of attack developed in (a):

r = d(P, C )

=√

(4 − 1)2 + (−1 − 2)2 distance formula

=√

32 + 32 = 3√

2

Thus, r = 3√

2. The equation of the circle is

(x − 1)2 + (y − 2)2 = (3√

2)2

(x − 1)2 + (y − 2)2 = 18


Solution to (c) As in the part (a), r = d(P, C), where C(−2, 3 ) andP ( 5, 2 ).

r = d(P, C)

=√

(5 − (−2))2 + (2 − 3)2

=√

49 + 1 =√

50

= 5√

2

The equation is for the desired circle is

(x + 2)2 + (y − 3)2 = (5√

2)2

or,(x + 2)2 + (y − 3)2 = 50

Exercise 10.8.


10.9. Solution: Since the two points P1 and P2 are diametricallyopposite, the radius is given by

r = 12 d(P1, P2 )

The center of the circle is the midpoint of the line segment P1P2;simple use the midpoint formula.


Solution to (b) P1( 1, 3 ) and P2( 3, 5 ).

I follow my own game plan:

r = 12d(P1, P2)

= 12

√(1 − 3)2 + (3 − 5)2

= 12

√8 = 1

2 (2√

2)

=√

2

The center is the midpoint between P1( 1, 3 ) and P2( 3, 5 )

C(h, k ) = C

(1 + 3

2,3 + 5

2

)

= C( 2, 4 )

The equation of the circle is

(x − 2)2 + (y − 4)2 = 2


Solution to (c) P1(−3, 1 ) and P2( 4, 6 )

r = 12

√(−3 − 4)2 + (1 − 6)2

= 12

√49 + 25 = 1

2

√74

The center is the midpoint between P1(−3, 1 ) and P2( 4, 6 ):

C(h, k ) = C

(−3 + 42

,1 + 6

2

)

= C( 1

2 , 72

)The equation of the circle is

(x − 12 )

2 + (y − 72 )

2 = 372

where we have calculated r2 = 14 (74) = 37

2 Exercise 10.9.


10.10. Solution to (a) x2 + y2 − 6x + 2y − 4 = 0

We complete the square and interpret the result.

x2 + y2 − 6x + 2y − 4 = 0 given

(x2 − 6x) + (y2 + 2y) = 4 group similar terms

(x2 − 6x + 9) + (y2 + 2y + 1) = 4 + 9 + 1 complete square

(x − 3)2 + (y + 1)2 = 14 center-radius form!

This last equation is in the center-radius form. It represents a circlewith center at C(3,−1) and radius r =

√14


Solution to (b) x2 + y2 + 4y = 2

x2 + y2 + 4y = 2 given

x2 + (y2 + 4y) = 2 group terms

x2 + (y2 + 4y + 4) = 6 complete square

x2 + (y + 2)2 = 6 center-radius form!

This is the equation of a circle with center at C(0,−2) and havingradius of r =

√6.


Solution to (c) 3x2 + 3y2 − 8x = 0

3x2 + 3y2 − 8x = 0 given

(3x2 − 8x) + 3y2 = 0 group similar terms

3(x2 − 83x) + 3y2 = 0 factor out leading coeff.

3(x2 − 83x + 16

9 ) + 3y2 = 163 complete square

3(x − 43 )

2 + 3y2 = 163 write as perfect square

(x − 43 )

2 + y2 = 169 center-radius form!

This is the equation of a circle with center at C( 43 , 0) and radius r = 4

3 .


Solution to (d) x2 + y2 + 2y − 4x + 10 = 0

x2 + y2 + 2y − 4x + 10 = 0 given

(x2 − 4x) + (y2 + 2y) = −10 group similar terms

(x2 − 4x + 4) + (y2 + 2y + 1) = −10 + 4 + 1 complete square

(x − 2)2 + (y + 1)2 = −5

The equation (x − 2)2 + (y + 1)2 = −5 does not describe a circlesince the number on the right-hand side is negative. There is no point(x, y) that satisfies this equation since the left-hand side is alwaysnonnegative and the right-hand side is negative. This equation has nograph. Exercise 10.10.


10.11. Solution to (a) Find the points of intersection between x2 +y2 = 4 and y = 2x. We equate the ordinates and solve for x. Theeasiest way of doing this is to substitute the value of y into the otherequation:

x2 + y2 = 4 and y = 2x

x2 + (2x)2 = 4 subst. y = 2x into circle

x2 + 4x2 = 4 expand

5x2 = 4 combine

x = ± 2√5

solve in one step!

These are the abscissas of intersection. To obtain the correspondingordinates, simple substitute these values into y = 2x to obtain thecoordinates of intersection:(

− 2√5,− 4√

5

) (2√5,

4√5

)


Solution to (b) Find the points of intersection between x2 + y2 = 4and y = x + 1.

x2 + (x + 1)2 = 4 substitute y = x+ 1 into first eq.

x2 + x2 + 2x + 1 = 4 expand

2x2 + 2x − 3 = 0 combine

Now apply the Quadratic formula:

x =−2 ± √

4 − 4(2)(−3)4

=−2 ± √

284

=−1 ± √

72

Summary : The abscissas of intersection are x =−1 ± √

72

.

The corresponding y-coordinates (the ordinates) can be obtained bysubstituting these abscissas into the straight line equation: y = x+1.I leave this to you.


Solution to (c) Find the points of intersection between x2+y2−2x = 4and y = 2x − 1.

x2 + (2x − 1)2 − 2x = 4 substitute in for y

x2 + 4x2 − 4x + 1 − 2x = 4 expand

5x2 − 6x − 3 = 0 combine

Now apply the Quadratic Formula:

x =6 ± √

36 − 4(5)(−3)10

=6 ± √

9610

=6 ± 4

√6

10

=3 ± 2

√6

5

Summary : The abscissas of intersection are x =3 ± 2

√6

5.


Solution to (d) Find the points of intersection between x2+y2−2x = 4and x + 2y = 6.

In this case, I think I’ll do something a little different. I’ll solve for xin the second equation, x = 6 − 2y and substitute this into the firstequation. This is to avoid working with fractions when it is really notnecessary to do so!

(6 − 2y)2 + y2 − 2(6 − 2y) = 4 Since x = 6− 2y

36 − 24y + 4y2 + y2 − 12 + 4y = 4 expand

5y2 − 20y + 20 = 0 combine

y2 − 4y + 4 = 0 simplify: Note perfect square!

(y − 2)2 = 0 write is so

y = 2 solve!

In this case, there is only one point of intersection:

y = 2 and x = 6 − 2y =⇒ x = 2Thus,


Point of Intersection: (2, 2)

If you draw the graphs of these two, the circle and the line, you willsee that the line is tangent to the circle at the point (2, 2).

C(1, 0)b

b

x+ 2y = 6(x − 1)2 + y2 = 5

Never mind, I did it myself! Exercise 10.11.


10.12. Solutions: The following answers are given without extensivediscussion. If you don’t see how I got the number, follow the tip givenabove.

(a) sin(3π/4) =√

2/2.

(b) cos(7π/6) = −√3/2. (Note:

7π6

= π +π

6.)

(c) sin(−7π/6) = − sin(7π/6) = −(−1/2) = 1/2. (Note: Same noteas previous note.)

(d) cos(−21π/2) = cos(21π/2) = 0. (Note:21π2

= 10π +π

2.)

(e) sin(9π/3) = sin(3π) = 0.

(f) cos(10π/3) = −1/2. (Note:10π3

= 3π +π

3.)

(g) sin(10π/3) = −√3/2. (Note: ditto.)


(h) cos(−29π/6) = cos(29π/6) = −√3/2. (Note:

29π6

= 5π − π

6;

i.e., 29π/6 is π/6 “short” of 5π. Wrap 5π, then come back π/6.)

Exercise Notes: I could have done part (h) in a way that is consistentwith how I did the others:

29π6

= 4π +5π6

,

Therefore,cos(29π/6) = cos(5π/6)

Now I would need to compute cos(5π/6). I found it much easier torealize that 29π/6 was π/6 “short” of 5π.

Though I didn’t refer to the property explicitly, what we are infact doing is using the periodic properties. In the previous remark,when I write

cos(29π/6) = cos(5π/6)


That is the periodic property of the cosine. To wrap around an amountof 29π/6 you first wrap 4π which is twice around the unit circle, thenyou go another 5π/6.

Exercise 10.12.


10.13. The values of x at which cos(x) takes on its minimum valueof y = −1 is

x = . . . , −5π, −3π, −π, π, 3π, 5π, . . .

In other words, cos(x) takes on its minimum value of y = −1 at allodd multiples of π. Exercise 10.13.


10.14. Solution: If you studied Table 1 and Figure 9, maybe youformulated the following answer:

x = . . . , −4π, −3π, −2π, π, 0, π, 2π, 3π, 4π, . . .

The function y = sin(x) is equal to zero at all integer multiples ofπ. Exercise 10.14.


10.15. Solution: From Table 1 and Figure 9 we see that

x = . . . , −13π2

, −9π2

, −5π2

, −π

2,

π

2,

5π2

,9π2

,13π2

, . . .

Every other odd positive multiple ofπ

2, and the negatives of all those

numbers.

Did I say a good english sentence? Let’s try again.

These numbers can be written in a more compact form:

x =π

2+ 2nπ n ∈ Z

That is, sin(x) = 1 for all numbers that differ fromπ

2by an integer

multiple of 2π.

Is that better?

Do you see the validity of these statement? Exercise 10.15.


10.16. Solution: From Table 1 and Figure 9 we see that

x = . . . , −15π2

, −11π2

, −7π2

, −3π2

,3π2

,7π2

,11π2

,15π2

, . . .

These numbers can be written in a more compact form:

x =3π2

+ 2nπ n ∈ Z

That is, sin(x) = −1 at all numbers that differ from3π2

by an integermultiple of 2π.

Do you see the validity of the statement? Exercise 10.16.


10.17. Solution: We have

cot(x) =cos(x)sin(x)

Therefore,

Dom(cot) = { x | sin(x) = 0 }= { x | x = nπ, n ∈ Z } from Exercise 10.14

Answer : Dom(cot) = { x | x = nπ, n ∈ Z } Exercise 10.17.


10.18. Answers:

(a)2π3

is the same as θ =2π3

180◦

π= 120◦.

(b) −5π6

is the same as θ = −5π6

180◦

π= −150◦

(c) 3π is the same as θ = 3π180◦

π= 540◦

(d) −3π4

is the same as θ = −3π4

180◦

π= −135◦

(e)5π2

is the same as θ =5π2

180◦

π= 450◦

Exercise 10.18.


10.19. Answers:

(a) 30◦ is the same as t = 30◦ π

180◦ =π

6.

(b) 60◦ is the same as t = 60◦ π

180◦ =π

3.

(c) 45◦ is the same as t = 45◦ π

180◦ =π

4.

(d) −300◦ is the same as t = −300◦ π

180◦ = −5π3

.

(e) 225◦ is the same as t = 225◦ π

180◦ =5π4

.

That seemed very easy. Exercise 10.19.


10.1. Solution: We begin by completing the square using the rulesgiven in Lesson 7.

y = x2 − 2x + 4 given, now associate x’s

y = (x2 − 2x) + 4 take 12 coef. of x and square it

y = (x2 − 2x + 1) + 4 − 1 add and subtract that

y − 3 = (x − 1)2 square completed!

The last equation has the form of (6); here, a = 1, h = 1 and k = 3.

This is a parabola that opens up and has vertex at V ( 1, 3 ).

Rather than accepting the rule for classification, let’s use this exampleto verify the assertion that V ( 1, 3 ) is the vertex.

The vertex is the point at which the parabola turns to create a maxi-mum or minimum. Because the coefficient of the square term, a = 1,


is positive, this vertex should represent a low point. We can see thiseasily from the equation

y − 3 = (x − 1)2

or,y = 3 + (x − 1)2

Notice that when x = 1, y = 3. The vertex is located at an valueof y = 3 on the vertical axis. Now, how does this compare with theordinate (the y-coordinate) of any other point on the graph? Observethat

y = 3 + (x − 1)2 ≥ 3,

because the expression (x − 1)2 ≥ 0. But this inequality states, atleast in my mind, that the ordinate, y, of a typical point on the graphis 3 or larger. Doesn’t this say that the vertex is the lowest point onon the graph? Example 10.1.


10.2. Solution: We simply complete the square! First though, weput it in the form of equation (6).

3 = 4x2 + 3x + 2y Given. Write as y = ax2 + bx+ c

y = 32 − 3

2x − 2x2 Done! Now complete square

y = −2(x2 + 34x) + 3

2 assoc. x’s and factor out coeff. of x

y = −2(x2 + 34x + 9

64 ) + 32 + 9

32 take 12 coeff. of x and square it

y = −2(x + 38 )

2 + 5732 done!

It is important that you fully understand each step—some detail wasleft out. In particular, make sure you understand the second to laststep above. Review the examples and exercises that follow the discus-sion of completion of the square.

Taking the constant in the last equation to the other side of the equa-tion we get

y − 5732 = −2(x + 3

8 )2


From this we can see that we have a parabola that opens down andwhose vertex is located at V (− 3

8 , 5732 ).

Be sure you understand why the first coordinate is h = − 38 and not

38 . (Compare with equation (6).) Example 10.2.


10.3. Solution: Simply apply the technique of completing the square.

Solution to (a) x2 + y2 − 2x + 4y + 1 = 0.

x2 + y2 − 2x + 4y + 1 = 0 given

(x2 − 2x) + (y2 + 4y) = −1 group similar terms

(x2 − 2x + 1) + (y2 + 4y + 4) = −1 + 1 + 4 complete sq. of each

(x − 1)2 + (y + 2)2 = 4 write as perfect squares

Summary : The center-radius form is

(x − 1)2 + (y + 2)2 = 4.

This is the equation of a circle with center at C( 1,−2 ) and radius ofr = 2.


Solution to (b) 2x2 + 2y2 + 4x − y − 2 = 0.

2x2 + 2y2 + 4x − y − 2 = 0 given

(2x2 + 4x) + (2y2 − y) = 2 group similar terms

2(x2 + 2x) + 2(y2 − 12y) = 2 factor out leading coeff.

(x2 + 2x) + (y2 − 12y) = 1 divide out these coeff.

(x2 + 2x + 1) + (y2 − 12y + 1

16 ) = 1 + 1 + 116 complete square

(x + 1)2 + (y − 14 )

2 = 3316 write as perfect squares

Summary : The equation written in the center-radius form is

(x + 1)2 + (y − 14 )

2 = 3316

This is the equation of a circle with center at C(−1, 14 ) and having

radius r =√

334 . Example 10.3.


10.4. All the requested values are for πn , for n = 1, 2, 3, 4, or 6, so

we can deduce the requested values using Table 1.

(a) Calculate sin(−π/3).

Solution: We can recall the odd property of the sine function

sin(−π/3) = − sin(π/3) = −√

32

by Table 1

Another way or reasoning is to realize that the ordinate terminalpoint P−π/3 is the negative of the ordinate of the terminal pointPπ/3 . . . this was the reasoning used to establish the symmetryproperties in the first place.

(b) Calculate cos(2π/3).

Solution: This is easy if you realize, from geometric considera-tions, that the terminal point of π/3 and 2π/3 are symmetri-cally placed on opposite of the y-axis. That being the case, thex-coordinates are opposite in sign; therefore,

cos(2π/3) = − cos(π/3) = −12

by Table 1


Depicted below are the points Pπ/3 and P2π/3; notice that Pπ/3 is“π/6” short of the vertical and P2π/3 is “π/6” beyond the vertical.This means they are symmetrical with respect to the vertical (they-axis).

bb

Pπ/3(x, y)P2π/3(−x, y)

• Question Verify the claim that, “Pπ/3 is ‘π/6’ short of the verticaland P2π/3 is ‘π/6’ beyond the vertical.”

(c) Calculate cos(5π/4).

Solution: Obviously,

5π4

= 54π = (1 1

4 )π = (1 + 14 )π = π +

π

4To wrap t = 5π/4 around the unit circle is equivalent to firstwrapping π units, and then wrap π/4 beyond that. The terminal


point, P5π/4 is in the third quadrant directly opposite the pointPπ/4.

b

b

Pπ/4(x, y)

P5π/4(−x, −y)

As you can see from the picture, both coordinates of P5π/4 arethe negative of their counterparts in the point Pπ/4; By Table 1we have cos(π/4) =

√2/2, we therefore have

cos(5π/4) = − cos(π/4) = −√

22

cos(5π/4) = −√

22


(d) Calculate sin(−π/2).

Solution: This one is easy,

sin(−π/2) = − sin(π/2) = −1 by Table 1

A picture is worth four words,

b

b

Pπ/2(0, 1)

P−π/2(0, −1)

sin(−π/2) = −1

Example 10.4.


10.5. Solution: Recall from equation (20) that we had identified allx at which cos(x) = 0 The cos(x) = 0 at all x such that

x = . . . , −5π2

, −3π2

, −π

2,

π

2,

3π2

,5π2

, . . .

This can be reformulated into a more compact form

x =π

2+ nπ n ∈ Z

That is, cos(x) = 0 all values of x that differ fromπ

2by an integer

multiple of π.

The domain of tan(x) is all x, except the ones listed above. Thus,

Dom(tan) ={

x | x = π

2+ nπ, n ∈ Z

}

Example 10.5.

Important Points

Quiz #1 on Parabolas

Quiz. Answer each of the following about the parabolas. (Some mayrequire some algebraic simplification.)1. Which of the following is a parabola defining y as a function of

x?(a) x + y − y2 = 2 (b) x + y − x2 = 2(c) xy = 3 (d) n.o.t.

2. Which of the following is a parabola defining y as a function ofx?(a) x + y + xy = 0 (b) x + y + x3 = 2(c) (x + y)2 = 1 (d) n.o.t.

3. Which of the following is a parabola defining y as a function ofx?(a) (x + 1)2 = (x + y) (b) (y + 1)2 = (x + y)(c) y(x + y) = 1 (d) n.o.t.

4. Which of the following is a parabola defining y as a function ofx?(a) 2x − 3y = 2 (b) (x − y)(x + y) = x(c) (x − y)(x + y) = 1 − y − y2(d) n.o.t.

EndQuiz. Important Point

Quiz #2 on Parabolas

Quiz. For each of the parabolas below, determine whether is opens upor down. Begin by putting the parabola in the form of a function ofx, then use equation (4).1. x + y − x2 = 2

(a) Opens Up (b) Opens Down2. x2 = x − y

(a) Opens Up (b) Opens Down3. (x + 1)2 = (x + y)

(a) Opens Up (b) Opens Down4. (x − y)(x + y) = 1 − y − y2

(a) Opens Up (b) Opens DownEndQuiz.

You will recognize these parabolas as the ones that appeared as theright answers in the Quiz in the section on recognition.

Important Point

Quiz #3 on Common Vales of Sines and Cosines

We reproduce the following equations for your convenience.

P0(1, 0) Pπ/2(0, 1) Pπ(−1, 0) P3π/2(0,−1) P2π(1, 0) (I-1)

The coordinates of these “easy” points contain the cosines and sinesof the corresponding values of t. (Given in the subscript of the point,I might note.)

Quiz. Using equation (I-1), answer each of the following correctly.Passing is 100%.1. cos(0) =

(a) −1 (b) 0 (c) 1 (d) n.o.t.2. sin(3π/2) =

(a) −1 (b) 0 (c) 1 (d) n.o.t.3. cos(π) =

(a) −1 (b) 0 (c) 1 (d) n.o.t.4. sin(π/2) =

(a) −1 (b) 0 (c) 1 (d) n.o.t.EndQuiz. Important Point

Quick Quiz

That’s right! The answer is Yes.

tan(x) =sin(x)cos(x)

sec(x) =1

cos(x)

Both of these functions are undefined when cos(x) = 0. In both cases

Dom(sec) = Dom(tan) = { x | cos(x) = 0}=

{x | x = π

2+ nπ, n ∈ Z

}

Important Point


Verify the claim that, “(1) Pπ/3 is ‘π/6’ short of the vertical and (2)P2π/3 is ‘π/6’ beyond the vertical.”

Verification of (1)π

2− π

6=

2π6

=π

3

Verification of (2)π

2+

π

6=

4π6

=2π3

. Important Point

algebrareviewintenlessons · 2011. 12. 28. · section 1: setting up the environment as any integer...

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