section 0.6: polynomial, rational, and …kkuniyuk.com/precalcbook/precalc0006to0010.pdfnumbers,...

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1

SECTION 0.6: POLYNOMIAL, RATIONAL, AND

ALGEBRAIC EXPRESSIONS

LEARNING OBJECTIVES

• Be able to identify polynomial, rational, and algebraic expressions.

• Understand terminology and notation for polynomials.

PART A: DISCUSSION

• In Chapters 1 and 2, we will discuss polynomial, rational, and algebraic

functions, as well as their graphs.

PART B: POLYNOMIALS

Let n be a nonnegative integer.

An n th -degree polynomial in x , written in descending powers of x, has the

following general form:

anxn+ an 1x

n 1+ ... + a1x + a0 , an 0( )

The coefficients, denoted by a

1, a

2,…, a

n, are typically assumed to be real

numbers, though some theorems will require integers or rational numbers.

an , the leading coefficient, must be nonzero, although any of the other

coefficients could be zero (i.e., their corresponding terms could be “missing”).

anxn is the leading term.

a0 is the constant term. It can be thought of as a

0x

0, where x

0= 1.

• Because n is a nonnegative integer, all of the exponents on x indicated above

must be nonnegative integers, as well. Each exponent is the degree of its

corresponding term.


Example 1 (A Polynomial)

4x35

2x2 +1 is a 3

rd-degree polynomial in x with leading coefficient 4,

leading term 4x3, and constant term 1. The same would be true even if the

terms were reordered:

15

2x2 + 4x3 .

The polynomial 4x35

2x2 +1 fits the form

anxn+ an 1x

n 1+ ... + a1x + a0 , with degree n = 3.

It can be rewritten as:

4x35

2x2 + 0x + 1 , which fits the form

a3x3+ a2x

2+ a1x + a0 , where the coefficients are:

a3= 4 leading coefficient( )

a2=

5

2

a1= 0

a0= 1 constant term( )

§

Example Set 2 (Constant Polynomials)

7 is a 0th-degree polynomial. It can be thought of as 7x

0.

0 is a polynomial with no degree. §


PART C: CATEGORIZING POLYNOMIALS BY DEGREE

Degree Type Examples

0 [Nonzero] Constant 7

1 Linear 3x + 4

2 Quadratic 5x2

x +1

3 Cubic x3+ 4x

4 Quartic x4

5 Quintic x5

PART D: CATEGORIZING POLYNOMIALS BY NUMBER OF TERMS

Number

of Terms Type Examples

1 Monomial x5

2 Binomial x3+ 4x

3 Trinomial 5x2

x +1

PART E: SQUARING BINOMIALS

Formulas for Squaring Binomials

a + b( )

2

= a2+ 2ab + b

2

a b( )

2

= a2 2ab + b

2

WARNING 1: When squaring binomials, don’t forget the “middle term” of the

resulting Perfect Square Trinomial (PST).

For example, x + 3( )2= x2 + 6x + 9 . Observe that 6x is twice the product of the

terms x and 3: 6x = 2 x( ) 3( ) .

The figure below implies that

x + y( )2

= x2+ 2xy + y

2 for x > 0 and y > 0 .


PART F: RATIONAL AND ALGEBRAIC EXPRESSIONS

A rational expression in x can be expressed in the form:

polynomial in x

nonzero polynomial in x

Example Set 3 (Rational Expressions)

Examples of rational expressions include:

a) 1

x.

b) 5x3 1

x2 + 7x 2. Irrational coefficients such as 2 are permissible

as coefficients of either polynomial.

c) x7+ x which equals

x7+ x

1. In fact, all polynomials are rational

expressions. §

An algebraic expression in x is also permitted to contain non-integer rational

powers of variable expressions (and their equivalents in radical form).

Example Set 4 (Algebraic Expressions)

Examples of algebraic expressions include:

a) x1/2

, or x .

b) x3 + 7x5/7

x x + 53+

.

• (See Footnote 1.)

All rational expressions are algebraic. §


The Venn diagram below is for expressions in x that correspond to functions

(see Chapter 1):

FOOTNOTES

1. Algebraic expressions. Some sources forbid the presence of in an algebraic expression,

since is a transcendental (i.e., non-algebraic) number. That means that is not a zero of

any polynomial with integer coefficients, as, say, 2 is.

(Section 0.7: Factoring Polynomials) 0.7.1

SECTION 0.7: FACTORING POLYNOMIALS

LEARNING OBJECTIVES

• Know techniques and formulas for factoring polynomials.

• Know the Test for Factorability for factoring quadratic trinomials.

• Recognize polynomials in quadratic form and be able to factor them.

PART A: DISCUSSION

• Factoring is a very commonly used technique in precalculus and calculus.

Factoring helps us simplify expressions, find zeros, solve equations and

inequalities, and find partial fraction decompositions (see Section 7.3).

• Rewriting a sum of terms as a product of factors helps us perform sign analyses,

as we will see in Sections 2.4 and 2.10.

PART B: FACTORING OUT GCFs

For now, when we factor a polynomial, we factor it completely over the integers

( ), meaning that the factors cannot be broken down further using only integer

coefficients. That is, the factors must be prime (or irreducible) over the integers.

• In Chapter 2, we will factor over other sets, such as

, , or .

TIP 1: The Greatest Common Factor (GCF), if it is not 1, should typically be

factored out first, although it can be factored out piece-by-piece for more

complicated expressions. (Unfortunately, there is no simple, standard definition for

the GCF.)

Example 1 (Factoring out a GCF)

We factor 8x + 6 as 2 4x + 3( ) , because 2 is the GCF.

2 is the greatest common divisor of 8 and 6. §

Example 2 (Factoring out a GCF)

We factor x5 + x3 as x3 x2 +1( ) , because x3 is the GCF.

x3 is the power of x with the least exponent. §


TIP 2: Sometimes, it is helpful to factor out 1, particularly when a polynomial

has a negative leading coefficient.

Example 3 (Factoring out -1 First)

Factor 8x5 6x

3 .

§ Solution

8x5 6x

3= 8x

5+ 6x

3( )= 2x

3 4x2+ 3( )

WARNING 1: Sometimes, people confuse signs if they try to

factor out 2x3 immediately.

§

WARNING 2: Be careful when factoring the base of a power. Make sure to apply

the exponent to all factors.

Example 4 (Factoring out of a Power)

x3 + x( )5 is not equivalent to x x2 +1( )

5.

The following is correct:

x3 + x( )5= x x2 +1( )

5

= x5 x2 +1( )5

Each factor of the base must be raised to the exponent, 5.

See Section 0.5, Law 4:

xy( )n

= xny

n . §


PART C: FACTORING FORMULAS

Factoring Formulas

Factoring a … Formula

Perfect Square Trinomial

(PST)

a2+ 2ab + b

2= a + b( )

2

a

2 2ab + b2= a b( )

2

Sum of Two Squares (A rule will be provided for a

2+ b

2

when we discuss imaginary numbers in

Section 2.1. As is, it is prime for now.)

Difference of Two Squares a

2b

2= a + b( ) a b( )

Sum of Two Cubes a3+ b

3= a + b( ) a

2ab + b

2( )

Difference of Two Cubes a

3b

3= a b( ) a

2+ ab + b

2( )

WARNING 3: Many math students forgot or never learned the last two formulas.

WARNING 4: In the last two formulas, there is no “2” or “ 2” coefficient on the

ab term of the trinomial factor. If a and b have no common factors (aside from 1

and 1), the trinomial factors are typically prime. Sometimes, people confuse

these trinomials with Perfect Square Trinomials (PSTs), which we introduced in

Section 0.6, Part E.

TIP 3: In the last two formulas, observe that the binomial factor is “as expected”:

a + b( ) for a3+ b

3, and a b( ) for a

3b

3. The visible signs on the right-hand

sides follow the pattern: “same,” “different,” and “+.”


PART D: TEST FOR FACTORABILITY and PRACTICE EXAMPLES

Test for Factorability

This test applies to any quadratic trinomial of the form ax2+ bx + c , where

a, b, and c are nonzero, integer coefficients.

(Assume the GCF is 1 or 1; if it is not, factor it out.)

The discriminant of the trinomial is b2 4ac .

• If the discriminant is a perfect square (such as 0, 1, 4, 9, etc.; these are

squares of integers), then the trinomial can be factored over the integers.

For example, x2+ 3x + 2 has discriminant 1 and can be factored as

x + 2( ) x +1( ) .

•• In fact, if the discriminant is 0, then the trinomial is a perfect

square trinomial (PST) and can be factored as the square of a

binomial with integer coefficients. For example, x2 6x + 9 has

discriminant 0 and can be factored as x 3( )2

.

• If the discriminant is not a perfect square, then the trinomial is prime

over the integers.

This test may be applied in Example Set 5, a) through g), which serve as

review exercises for the reader.

• The discriminant is denoted by (uppercase delta), though that symbol is also used

for other purposes. It is seen in the Quadratic Formula in Section 0.11. We will discuss a

method for factoring quadratic trinomials using the Quadratic Formula in Chapter 2.


Example Set 5 (Factoring Polynomials)

Factor the following polynomials over the integers.

a) x2+ 9x + 20

b) x2 20x +100 (Hint: This is a Perfect Square Trinomial (PST).)

c) x2 4x 12

d) 3x2 20x 7

e) 4x2+11x + 6

f) 2x2+10x + 5

g) 3x2+ 6x 3

h) x4 16

i) a3 3a + 2a

2b 6b

(Hint: Use Factoring by Grouping. This is when we group terms and factor

each group “locally” before we factor the entire expression “globally” by

factoring out the GCF.)

j) 4x2+ 9y

2

k) x3+125y

3

l) x3 125y

3


§ Solution

a) ax2 + bx + c = x2 + 9x + 20 = x + 5( ) x + 4( )

We want 5 and 4, because they have product = c = 20 and (since this is the

a = 1 case) sum = b = 9. We can rearrange the factors:

x + 4( ) x + 5( ) .

b)

x2

x( )2

20x + 100

10( )2

Guess that this is aPST for now.

= x 10( )

Check:2 x( ) 10( )=

20x

2, or x 10( ) x 10( ) = x 10( )

2

c) x2 4x 12 = x 6( ) x + 2( )

How do we know we need 6 and + 2?

The constant term, c, is negative, so use opposite signs: one "+" and one " ."

The middle coefficient, b, is negative, so the negative number must be higher in

absolute value than the positive number; it “carries more weight.”

2-factorizations of –12 (which is c)

Think: What? • What?? = –12.

Sum = b = 4?

( a = 1 case )

12, +1 No

6, +2 Yes Can stop

4, +3 No

d) F + O + I( ) + L = 3x2 20x 7 = 3x + 1( ) x 7( )

F = First product (product of the First terms)

O = Outer product (product of the Outer terms)

I = Inner product (product of the Inner terms)

L = Last product (product of the Last terms)

3x( ) x( ) F = 3x2 ; factors must be 3x and x

Need L = 7

+ 7 1

1 + 7 Makes O + I = 20x. We need O + I to be 20x,

which is the middle term of the trinomial.

We're only off by a sign, so we change both

signs.

+ 1 7 Makes O + I = 20x. This works.

7 + 1

Also, b = 20 , a "very negative" coefficient, so we are inclined to pair up the 3x

and the 7 to form the outer product, since they form 21x .


e) 4x

2+11x + 6 = 4x + 3( ) x + 2( )

Method 1: Trial-and-Error ("Guess") Method

( )( )

4x x

2x 2xF = 4x2

+ 1 + 6

+ 6 + 1

+ 2 + 3

+ 3 + 2

L = 6; need both " + " because of + 11x

Method 2: Factoring by Grouping

4 and 6 are neither prime nor “1,” so we may prefer this method.

We want two integers whose product is ac = (4)(6) = 24

and whose sum is b = 11. We want 8 and 3; split the middle

term accordingly.

4x2+ 11x + 6 = 4x2

+ 8x + 3x

OK to switch

+ 6

= 4x2+ 8x( ) + 3x + 6( ) Group terms

= 4x x + 2( ) + 3 x + 2( ) "Local factoring"

= 4x + 3( ) x + 2( ) "Global factoring"

f) 2x2+10x + 5 is prime or irreducible over the integers (i.e., it cannot be broken down

further using integer coefficients). None of these combinations work:

2x( ) x( ) F = 2x2 ; factors must be 2x and x

Need L = 5; need both " + " because of + 10x

+ 1 + 5

+ 5 + 1

We could also apply the Test for Factorability. The discriminant

b2 4ac = 10( )2

4 2( ) 5( ) = 100 40 = 60 , which is not a perfect square, and

the GCF = 1, so the polynomial is prime.

g)

3x2+ 6x 3 = 3

GCF

x2 2x + 1( )a PST

You should usually factor out the GCF first.

= 3 x 1( )2


h) Apply the Difference of Two Squares formula a2 b2 = a + b( ) a b( )[ ] twice:

x4

x2( )

2

16

4( )2

= x2+ 4( )

prime

x2

x( )2

4

2( )2

= x2+ 4( ) x + 2( ) x 2( )

i) Use Factoring by Grouping:

a3 3a + 2a

2b 6b = a

3 3a( ) + 2a2b 6b( )

= a a2 3( ) + 2b a

2 3( )= a + 2b( ) a

2 3( )

j) 4x2+ 9y

2 is prime. The GCF = 1, and we have no formula for the Sum of Two Squares

(for now…; this will change when we discuss imaginary numbers in Section 2.1).

k) Apply the Sum of Two Cubes formula

a3+ b3

= a + b( )

"Expectedfactor"

a2 ab

NOT2ab

+ b2

The visible signs follow the pattern:same, different, "+"

:

x3

x( )3

+ 125y3

5y( )3

= x + 5y( ) x2 5xy + 25y2( )

l) Apply the Difference of Two Cubes formula

a3 b3= a b( )

"Expectedfactor"

a2+ ab

NOT+2ab

+ b2

The visible signs follow the pattern:same, different, "+"

:

x3

x( )3

125y3

5y( )3

= x 5y( ) x2 + 5xy + 25y2( ) . §


PART E: FACTORING EXPRESSIONS IN QUADRATIC FORM

An expression is in quadratic form

It can be expressed as au2+ bu + c after performing a u substitution, where

a 0 , and a, b, and c are real coefficients.

• The term “quadratic form” is defined differently in higher math; that definition requires

each term to have degree 2.

Example 6 (Factoring an Expression in Quadratic Form)

Factor 2x6 x3 1 over the integers.

§ Solution

The trinomial 2x6 x3 1 is in quadratic form, because the exponent on x

in the first term is twice that in the second term (6 is twice 3), and the third

term is a constant.

We will use the substitution u = x3, the power of x in the “middle” term.

Then, u

2= x

3( )2

= x6 .

2x6 x3 1= 2u2 u 1

Now, factor as usual.

= 2u +1( ) u 1( )

Substitute back. Replace u with x3.

= 2x3+1( ) x3 1( )

With practice, the substitution process can be avoided. Either way, we are

not done yet! It is true that 2x3 +1( ) is prime over the integers; Chapter 2

will help us verify that. However, x3 1( ) is not prime, because we can

apply the Difference of Two Cubes formula.

2x3 +1( ) x3 1( ) = 2x3 +1( ) x 1( ) x2 + x +1( )

This is factored completely over the integers. The Test for Factorability

can be used to show that the trinomial factor x2 + x +1( ) is prime, as

expected. §

(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.1

SECTION 0.8: FACTORING RATIONAL AND ALGEBRAIC

EXPRESSIONS

LEARNING OBJECTIVES

• Know techniques for factoring rational and algebraic expressions.

PART A: DISCUSSION

• In this section, we will extend techniques for factoring polynomials to other

rational and algebraic expressions, including those with negative and fractional

exponents. Prior to a precalculus course, most students have no experience

factoring such expressions.

PART B: FACTORING OUT GCFs

When we factor x5 + x3 as x3 x2 +1( ) , we factor out x3, the power of x with the

least exponent; x3 is the GCF. We then divide each term of x5 + x3 by x

3 to

obtain the other factor, x2 +1( ) . When we divide x5 by x3 , we subtract the

exponents in that order and get x2 .

TIP 1: Think: “We’re factoring x3 out of x5 . 5 takeaway 3 is 2.”

These techniques apply even when the exponents involved are negative and/or

fractional.


Example 1 (Factoring with Negative Exponents)

Factor x 7+ x 4 2x 1

over the integers.

§ Solution

Observe that 7 is the least exponent on x. Our GCF is x7, so we will

factor it out and subtract 7 from each of the exponents.

x 7+ x 4 2x 1

= x 7 1+ x 4 7( ) 2x 1 7( )( )= x 7 1+ x 4+ 7 2x 1+ 7( )

= x 7 1+ x3 2x6( )

TIP 2: Observe that this last trinomial has no

negative exponents on x. This is a sign that we

have factored out the GCF correctly.

WARNING 1: We usually try to avoid negative

exponents in final answers, so we will rewrite the

expression as a fraction.

=1+ x3 2x6

x7

We are not done yet! We can factor the numerator

further over the integers.

TIP 3: We will first factor out 1 so that the new

leading coefficient is positive. This tends to make

factoring easier.

=1 x3 + 2x6( )

x7

The indicated 1 factor can be moved in front of

the fraction with no other sign changes.

WARNING 2: This is because it is a factor of the

entire numerator.

=1 x3 + 2x6

x7


TIP 4: We will now rewrite the numerator in

descending powers of x. This tends to make

factoring easier.

=2x6 x3 1

x7

Fortunately, we have already factored the

numerator in Section 0.7, Example 6.

=2x3 +1( ) x3 1( )

x7

=2x3 +1( ) x 1( ) x2 + x +1( )

x7

§

Example 2 (Factoring with Negative and Fractional Exponents)

Factor x3 + 2( )1/3

+ x3 + 2( )5/3

over the integers.

WARNING 3: Exponents do not typically distribute over sums.

WARNING 4: Likewise, the root of a sum is not typically equal to the sum

of the roots.

§ Solution

WARNING 5: All negative exponents are less than all positive exponents.

Observe that 5

3 is the least exponent on x3 + 2( ) . Our GCF is x3 + 2( )

5/3,

so we will factor it out and subtract

5

3 from each of the exponents on

x3 + 2( ) .


x3 + 2( )1/3

+ x3 + 2( )5/3

= x3 + 2( )5/3

x3 + 2( )1

3

5

3 + 1

= x3 + 2( )5/3

x3 + 2( )1

3+5

3 + 1

= x3 + 2( )5/3

x3 + 2( )2+ 1

= x3 + 2( )5/3

x6 + 4x3 + 4 + 1

= x3 + 2( )5/3

x6 + 4x3 + 5

=x6 + 4x3 + 5

x3 + 2( )5/3

Although x6 + 4x3 + 5 is in quadratic form, it is

prime over the integers. §

(Section 0.9: Simplifying Algebraic Expressions) 0.9.1

SECTION 0.9: SIMPLIFYING ALGEBRAIC EXPRESSIONS

LEARNING OBJECTIVES

• Be able to use factoring and canceling (i.e., division) to simplify algebraic

expressions.

• Be able to rationalize the numerator or the denominator of a fraction

containing radical(s).

• Be aware of common errors when rewriting and simplifying expressions.

PART A: DISCUSSION

• The factoring techniques from Sections 0.7 and 0.8 will help us simplify fractions

by canceling (i.e., dividing out) common factors.

• We can also re-express or simplify a fraction by rationalizing the numerator or

the denominator.

PART B: CANCELING COMMON FACTORS IN A FRACTION

We are typically required to provide answers in simplified form. While there are

different opinions as to what that means, there is common agreement that factors

(other than 1 and 1) that are common to both the numerator and the denominator

of a fraction must be canceled (i.e., divided out).

WARNING 1: Some instructors object to the term “canceling,” because students

often abuse the idea by inappropriately and carelessly deleting matching

expressions. Remember to rely on mathematical rules, not merely wishful

thinking!

Canceling Rule for Fractions

As we simplify a fraction, we can “cancel”:

a nonzero factor of the entire numerator with

an equivalent factor of the entire denominator.

• That is, we can divide the entire numerator and the entire denominator

by equivalent nonzero factors.


For example, we can divide the numerator and the denominator of

x

x2

x + 4( )

by x and obtain

1

x x + 4( ). We have canceled a pair of x factors:

x

1( )

x2

x( )x + 4( )

=1

x x + 4( )

WARNING 2: We cannot cancel a pair of x factors in x

x2 + 4, because x is

not shown as a factor of the entire denominator.

Canceling and Restrictions

When canceling, if a factor is eliminated from the denominator

of a fraction, check to see if any restrictions have been “hidden.”

Such restrictions must be written out separately.

• It is usually acceptable to wait until the end to write such restrictions.

For example,

x2

x x + 4( )=

x2

x( )

x

1( )x + 4( )

x 0( )

=x

x + 4x 0( )


PART C: CANCELING OPPOSITE FACTORS

a b( ) and b a( ) are opposites.

Canceling Opposite Factors

The quotient of nonzero, opposite factors is 1.

Example 1 (Canceling Opposite Factors)

Simplify

9 x2

x 3.

§ Solution Method 1

9 x2

x 3=

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

WARNING 3: It is incorrect to factor 9 x2 as

x + 3( ) x 3( ) ,

which is the factorization of x2 9 .

3 x( ) and

x 3( ) are opposites, so their quotient in either

order is 1.

=3+ x( ) 3 x( )

1( )

x 3( )1( )

x 3( )

TIP 1: The parentheses around the “ 1” remind us that it is a

factor of the numerator, not a term.

= 3+ x( ) x 3( ) , or

3 x x 3( )

The choice between

3+ x( ) and 3 x may depend on context. §


§ Solution Method 2

9 x2

x 3=

x2 9( )

x 3

=x

2 9

x 3

=x + 3( ) x 3( )

1( )

x 3( )1( )

x 3( )

= x + 3( ) x 3( ) , or

x 3 x 3( )

§

PART D: SIMPLIFYING ALGEBRAIC EXPRESSIONS

Example 2 (Simplifying an Algebraic Expression)

Simplify

4x + 7( )1/32x( ) x2( )

134x + 7( )

2/34( )

4x + 7( )2/3 .

Do not leave nonpositive exponents in the final expression.

• In calculus, the given expression is obtained by applying the Quotient Rule for

Differentiation to x2

4x + 73.

§ Solution

We begin by “cleaning up” the numerator.

4x + 7( )1/32x( ) x2( )

134x + 7( )

2/34( )

4x + 7( )2/3

=

2x 4x + 7( )1/3 4

3x2 4x + 7( )

2/3

4x + 7( )2/3


Method 1 (Factor the Numerator)

We will factor out the GCF of the numerator, which is x 4x + 7( )2/3

.

TIP 2: It may be easier to factor out x in one step and then

4x + 7( )2/3

in another step. The GCF does not have to be factored

out in one step.

TIP 3: It may be easier to substitute u = 4x + 7 and obtain

2xu1/343x2u 2/3

u2/3. Make sure to substitute back later.

2x 4x + 7( )1/3 4

3x2 4x + 7( )

2/3

4x + 7( )2/3 or

2xu1/3 43x2u 2/3

u2/3

=

x 4x + 7( )2/3

2 4x + 7( )1

3

2

343x

4x + 7( )2/3 or

xu 2/3 2u1

3

2

3 43x

u2/3

=

x 4x + 7( )2/3

2 4x + 7( )43x

4x + 7( )2/3 or

xu 2/3 2u43x

u2/3

We now divide 4x + 7( )2/3

by 4x + 7( )2/3

.

If we let u = 4x + 7 :

4x + 7( )2/3

4x + 7( )2/3 =

u 2/3

u2/3= u

2

3

2

3 = u4

3 =1

u4 /3=

1

4x + 7( )4 /3

The division yields 4x + 7( )4 /3

in the denominator.

=

x 2 4x + 7( )43x

4x + 7( )4 /3

=

x 8x +1443x

4x + 7( )4 /3


WARNING 4: Compound fractions are typically unacceptable in a

simplified expression.

We will multiply the numerator and the denominator by 3.

=

3x 8x +1443x

3 4x + 7( )4 /3

=24x2

+ 42x 4x2

3 4x + 7( )4 /3

=20x2

+ 42x

3 4x + 7( )4 /3 , or

2x 10x + 21( )

3 4x + 7( )4 /3

WARNING 5: We should factor the numerator and see if there are

any common factors with the denominator, aside from 1 and 1.

Such factors must be canceled (divided out).

WARNING 6: Do not distribute the 3 in the denominator. The 4/3

exponent forbids that.

• No restrictions need to be written. The only restriction on the given

expression was

x7

4, and it is implied by the final expression.

Method 2 (Rewrite as a Compound Fraction First)

Because of the negative exponent 2

3, the expressions in blue below can be

rewritten as fractions containing positive exponents.

Again, the substitution u = 4x + 7 may help.

2x 4x + 7( )1/3 4

3x2 4x + 7( )

2/3

4x + 7( )2/3 or

2xu1/3 43x2u 2/3

u2/3

=

2x 4x + 7( )1/3 4x2

3 4x + 7( )2/3

4x + 7( )2/3 or

2xu1/3 4x2

3u2/3

u2/3


We will multiply the numerator and the denominator by the Least

Common Denominator (LCD), 3 4x + 7( )2/3

, or 3u2/3 .

=

3 4x + 7( )2/3

2x 4x + 7( )1/3 4x2

3 4x + 7( )2/3

3 4x + 7( )2/3

4x + 7( )2/3

or

3u2/3 2xu1/3 4x2

3u2/3

3u2/3 u2/3

WARNING 7: Do not cancel the expressions above in red. We must

apply the Distributive Property. If it helps, write the following step;

with experience, you can skip it.

=

3 4x + 7( )2/3

2x 4x + 7( )1/3

3 4x + 7( )2/3 4x2

3 4x + 7( )2/3

3 4x + 7( )2/3

4x + 7( )2/3

or

3u2/3 2xu1/3 3u2/3 4x2

3u2/3

3u2/3 u2/3

=6xu 4x2

3u4 /3

=6x 4x + 7( ) 4x2

3 4x + 7( )4 /3

=24x2

+ 42x 4x2

3 4x + 7( )4 /3

=20x2

+ 42x

3 4x + 7( )4 /3 , or

2x 10x + 21( )

3 4x + 7( )4 /3

§


PART E: RATIONALIZING DENOMINATORS AND NUMERATORS

When we rationalize the denominator of a fraction, we eliminate radicals and

non-integer exponents in the denominator but possibly introduce them in the

numerator.

For example,

1

2 is unacceptable in a simplified expression, so we

rewrite it:

1

2=

1

2

2

2=

2

2. For more complicated expressions, there

are different opinions as to when denominators need to be rationalized. (See Footnote 1)

When we rationalize the numerator of a fraction, we eliminate radicals and

non-integer exponents in the numerator but possibly introduce them in the

denominator.

Example 3 (Rationalizing a Numerator)

Re-express

x + h x

h by rationalizing the numerator.

• This expression is an example of a difference quotient (see Section 1.10).

Rationalizing the numerator will help us find a derivative (see Section

1.11).

§ Solution

We will multiply the numerator and the denominator by x + h + x ,

the conjugate of x + h x .

x + h x

h=

x + h x( )h

x + h + x( )x + h + x( )

To multiply out the numerator, we use the

rule: a b( ) a + b( ) = a

2b

2 .


=x + h( )

2

x( )2

h x + h + x( )

WARNING 8: Do not forget the radical

expression in the denominator.

=x + h( ) x

h x + h + x( )

=x + h x

h x + h + x( )

=h

h x + h + x( )

We now cancel the h factors and note the

restriction h 0( ) .

=h

1( )

h

1( )x + h + x( )

h 0( )

=1

x + h + x

h 0( )

§


PART F: COMMON ERRORS

Beware of the following errors when rewriting or simplifying an expression.

Errors involving Radicals and Exponents

Error (crossed out) Related Correct Formulas Comments

a

2= a, if a < 0

a2= a

a33= a

a( )2

= a, if a 0

The formula a

2= a is

often forgotten.

a + b = a + b

a b = a b

a + b( )1/2

= a1/2

+ b1/2

a b( )1/2

= a1/2

b1/2

ab = a b ,

if a 0 and b 0 .

a

b=

a

b

,

if a 0 and b > 0 .

Do not apply radicals or

exponents term-by-term.

There is no general

formula for the square

root of a sum (or a

difference), as there is

for a product or a

quotient.

ab + ac = a b + c

ab + ac( )

1/2

= a b + c( )1/2

ab + ac = a b + c( )

= a b + c ,

if a 0 and b + c 0.

ab + ac( )1/2

= a b + c( )1/2

= a1/2

b + c( )1/2

,

if a 0 and b + c 0.

When factoring a

radicand or the base of a

power, make sure to

apply the appropriate

radical or exponent to all

factors.


Errors involving Fractions

Error (crossed out) Related Correct Formulas Comments

x1( )

x2x( )+ 4

=1

x + 4

x2

x x + 4( )=

x2

x( )

x

1( )x + 4( )

x 0( )

=x

x + 4x 0( )

We can cancel (nonzero,

equivalent) factors of the

entire numerator and the

entire denominator.

1

a+

1

b=

1

a + b

a

1+

b

1=

a + b

1.

1

a+

1

b=

1

a

b

b+

1

b

a

a

=b + a

ab, or

a + b

ab.

If we add like fractions,

we add the numerators

and keep the common

denominator.

If we add unlike

fractions, we do not add

the denominators.

a

d

b + c

d=a b + c

d

a

d

b + c

d=a b + c( )

d

A fraction groups

together its numerator

(and groups together its

denominator).

1

x + x2= x

1+ x

2

1

x+

1

x2= x

1+ x

2 Reciprocals cannot be

taken term-by-term.

1

7x 4 /3 = 7x4 /3

1

7x 4 /3 =1

7x4 /3 x 0( )

The negative exponent

only applies to x.

FOOTNOTES

1. Rationalizing denominators. The expression 2x 10x + 21( )

3 4x + 7( )4 /3 from Example 2 is often

considered simplified. If we were to rationalize the denominator, we would have (nontrivial)

powers of 4x + 7( ) as factors of the numerator and the denominator:

2x 10x + 21( )

3 4x + 7( )4 /3 =

2x 10x + 21( )

3 4x + 7( )4 /3

4x + 7( )2 /3

4x + 7( )2 /3 =

2x 10x + 21( ) 4x + 7( )2 /3

3 4x + 7( )2

We may be inclined to divide those powers and go back to 2x 10x + 21( )

3 4x + 7( )4 /3 .

(Section 0.10: More Algebraic Manipulations) 0.10.1

SECTION 0.10: MORE ALGEBRAIC MANIPULATIONS

LEARNING OBJECTIVES

• Learn additional ways to write equivalent expressions.

PART A: DISCUSSION

• The techniques of this section will help us rewrite expressions so that they are

easier to manipulate.

• They will also help us fit expressions to particular forms and templates.

PART B: SPLITTING A FRACTION THROUGH ITS NUMERATOR

Example 1 (Splitting a Fraction Through its Numerator)

Fill in the boxes below with appropriate, simplified numbers:

7x + 3

x4= x + x

§ Solution

We begin by rewriting x4

as x1/4

, because the template on the right-hand

side suggests a preference for exponential form over radical form.

7x + 3

x4=7x + 3

x1/4

We will rewrite this as a sum by splitting it through its

numerator.

=7x1

x1/4+3

x1/4

= 7x3/4 + 3x 1/4


Answer: 7x + 3

x4= 7 x

3

4+ 3 x

1

4

• In calculus, the right-hand side is much easier to differentiate and integrate.

§

WARNING 1: Do not split a fraction in this way through its denominator.

For example, 1

x + y is not equivalent to

1

x+1

y.

PART C: MAKING COMPOUND FRACTIONS

When we divide by a nonzero number, we multiply by its reciprocal.

When we multiply by a nonzero number, we divide by its reciprocal.

Example 2 (Making Compound Fractions)

Fill in the boxes below with appropriate, simplified numbers:

9x2 +3y2

4=

x2+

y2

§ Solution

9x2 +3y2

4=9

1x2 +

3

4y2

WARNING 2: 3

4y2 is not equivalent to

3

4y2 .

=x2

19

+y2

43

Answer: 9x2+

3y2

4=

x2

19

+y2

43

§

• This technique will be used in Chapter 10 to set up standard forms for equations of conics.


PART D: COMPENSATION

Sometimes, we can alter an expression if we compensate for the change and

maintain equivalence.

Example 3 (Compensation Using Addition and Subtraction)

Fill in the box below with an appropriate, simplified expression:

x

x +1= 1

§ Solution

We know that x +1

x +1= 1 x 1( ) , so we would like to add 1 to the

numerator of x

x +1. To compensate for that, we must also subtract 1 from

the numerator.

WARNING 3: If we add 1 to the numerator of a fraction, we cannot

compensate by adding 1 to the denominator. For example,

1

2

2

3.

x

x +1=x +1 1

x +1

=x +1

x +1

1

x +1

= 11

x +1

Answer: x

x +1= 1

1

x +1

§

• A similar compensation technique is used when Completing the Square (CTS) in Sections 0.11,

0.13, and 2.2 and Chapter 10.


Example 4 (Compensation Using Multiplication and Division)

Fill in the box below with an appropriate, simplified number:

4x +1( )5= 4x +1( )

54( )

§ Solution

The expression on the left-hand side has been multiplied by 4.

We compensate by dividing by 4, or by multiplying by its reciprocal, 1

4.

Answer: 4x +1( )5=

1

44x +1( )

54( )

• In calculus, this technique helps us integrate using the u-substitution method.

section 0.6: polynomial, rational, and …kkuniyuk.com/precalcbook/precalc0006to0010.pdfnumbers,...

Documents