chapter 2 (math 10-3)

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Chapter 2

ALGEBRAIC EXPRESSIONS

Every day, we encounter situations or phenomena that we need to study and analyze.

Expressing this mathematically will simplify the whole process of analysis. The mathematical

expressions used to model situations may involve quantities of known values or quantities whosevalues are changing.

This chapter introduces the concept of algebraic expressions. It also includes a discussion

of the terminologies and principles, as well as the operations, which apply to algebraicexpressions.

2.1 ALGEBRAIC EXPRESSIONS

Mathematical statements use two types of quantities: constant and variables. A variable

is a quantity whose value changes while a constant is a quantity whose value does not change. It

is common to represent each of these quantities by a symbol and since the value of a constant

does not change, it is normal to use in place of the symbol the value itself. The following areexamples of constants:

ea f) 3.1375- e) 5 d) c) 2

1 b) 2 )

On the other hand, since a variable takes on different values, it is customary to represent

it by a symbol. The values that the variables can be substituted with can be any real number and

as such the properties of the real numbers apply for variables and constants. By convention, anyletter of the English or Greek alphabet is used as a variable name. The following are examples of

variables:

a) t b) s c)u d) e) f)

When constants and variables are associated by any of the fundamental operations, we

have an algebraic expression. By definition,

The following are examples of algebraic expressions:

a) 2 x c) y

2 e) x 2 y2 , 3 x2y-2xy2 3y3 g ) 3x

b) x y 3 d ) 2 x4y

f ) xy

An algebraic expression is the result of associating constants and

variables by addition, subtraction, multiplication, division, including roots and

powers.

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In the expression x2 y, each of the factors 2 , y and x can be considered as a coefficient

of the other, where 2 is called the numerical coeff icient of x y and xy is the li teral coeff icient of2. The expression consisting of a product of constants and variables including the (+) or ( – ) sign

preceding it is called an algebraic term or simply a term. Thus in the expression:

a) x2 there is one term

b) 5(x+y) there is one term

c) 2 x2 4 y( xy1) there are 2 terms, namely: )1(4,22 xy y x

d) 3y2xy-y3 322 x there are 3 terms, namely: 3y,2xy-y,3 322 x

In expressions such as xy- andy3 , it is implied that the numerical coefficients are 1 and

-1, respectively. When two or more terms have the same literal coefficients, they are called

similar or like terms.

An algebraic expression may involve a power given by the formna , where a is called

the base and n , the exponent. The power expressionna is a product of n-factors of a.

2.2 TYPES OF ALGEBRAIC EXPRESSIONS

An algebraic expression may be classified as a polynomial , a rational expression or an

ir rational expression .

The following are examples of polynomials.

475362 ) 23456 x x x x x xa

96 ) 48 x xb 6542332456

61520156 ) y xy y x y x y x y x xc 1 ) 10 xd

Polynomials may be classified according to degree, according to the number of terms present, or according to the nature of the numerical coefficient.

A polynomial is an algebraic expression consisting of one or more terms inwhich the exponent of the variable or of the variables is a non-negative integer. A polynomial may involve one or more variables.

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The degree of a term is the sum of the integral exponents of the variables in the term.

Thus, the degree of each of the following terms is as indicated.

8 )a zero degree

wb 7 ) 1st degree

62- ) xc 6th degree227 ) y xd 4th degree

37 ) yz xe 11th

degree

The degree of a polynomial in one variable is the highest degree of the variable in the

polynomial. If the polynomial is in two or more variables, then the degree of the polynomial is

the highest degree among the terms of the polynomials. Polynomials are classified according todegree as:

a) zero degree

b)

1st degree or linearc) 2

nd degree or quadratic

d) 3rd

degree or cubic

e) 4th

degree or quartic f) nth degree , for any n a positive integer

The following polynomials are classified according to its degree.

23456 5362 ) x x x x xa 6th degree

43- ) 14511 x x xb 14th degree

y x xy y xc 435 23 ) 6th degreew-x6w27x- ) 4224 d 4th degree

12 ) xe 1st degree or linear

When polynomials are classified according to the number of terms it has, the polynomial

may be described as:

a) monomial – a polynomial of one term

b) binomial - a polynomial of two termsc) trinomial – a polynomial of three terms

d)

multinomial or simply polynomial – polynomial of four or more terms

Thus;62 ) xa monomial

47 ) xb binomial

23 ) 35 x xc trinomial

4753 ) 23 x x xd multinomial or polynomial

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Polynomials can also be classified according to the nature of its numerical coefficient as

being integral, rational or i rr ational . A polynomial that is integral has integers as numerical

coefficients of all of the terms. When some of the numerical coefficients are expressed as a ratioof two integers or as a fraction or as decimal numbers, the polynomial is said to be rational. At

times, some of the coefficients of the terms are irrational numbers. In cases like these, the

polynomial is said to be a real polynomial. Thus an integral or a rational polynomial is also a real polynomial. The following illustrates this concept.

475362 ) 23456 x x x x x xa integral, real

x x xb 24 ) 38 integral, real

12 ) 53 y yc real1 ) xd real

e)1

2 x2

3

4 x 4 rational, real

A rati onal expression is an algebraic expression involving a ratio of two polynomials.Examples are the following:

2x

3 a)

2x

y-x b)

42x

1x c)4

3

x

An ir rational expression is an algebraic expression that involves variables raised to

fractional exponents, such as the following:

12xy ) a 3zy-2x-xyz ) 3 b y-x ) 4 3c

2

1

2

1

) y xd uv

w )

5

4

e

Exercises:

I. Classify each of the following as to the nature of the algebraic expression.

1. 7 x4 2 x 4 x2 3

2. 5 x x 2

3 x4 16 x2

3. 2 yx2 xy2 3 x3 5 y3 1

4.1

2 x 2 x23 x3 5

5. x

y2 x 7 y2 2 y

2

x

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II. Complete the table below by classifying each of the given polynomials according

to the stated parameters.

Classification according to:Polynomial Number ofterms

Degree Nature of Numerical

Coefficient

1. 8 2 x 2. x6 3 x2 2 x 3

3.1

3( y 3)

2

5( y 3)7 1

4. 2 x x2 ex 3 5

5. x2 y 3 xy2 3 x3 4 y4 1 6. ab2c3 2a3bc2 4a3b2c a2b2c2

7. 52mn4 3m2n3 m3n2 3m4n 2m5 n5

EVALUATION OF ALGEBRAIC EXPRESSIONS

The value of an algebraic expression can be evaluated at particular values of the variables

by substituting the given values in the expressions and evaluating the expression subject to the

rules of signed numbers. The value of each of the following expressions is as follows:

a) 362 22 xy x y x when evaluated at x = 2 and y = 3 is

3)3)(2()2(6)3()2(2

22

= 27

b)42x

1x 4

3

x when evaluated at x = -1 is

4)1(2(-1)

1(-1) 4

3

=

421

11-

=

5

2

=

5

2

c) 1 x when evaluated at x = 5 is

1)5( = 15

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Exercises: Evaluate each of the following algebraic expressions at the indicated values of

the variables.

1. ))(( 33 nmnm where m = 2 and n = -1

2. 22 23

)2)((

y xy x

y x y x

where x = 3 and y = 1

3.2)4(3

2 y y x

x where x = 2 and y = 4

4.4224 2 s sr r where r = 4 and s = 2

5.222222 ya xbba

ab

c where a =4, b = 9, c = 1, x=2, and y = 1

6. ( x 2)2 y 3( x 2) y2 3( x 2)3 4 y4 1 where x = 1 and y = -2

7. 2 x2 y3 x

y2 xy 4 where x = 2 and y = 1

2.3 INTEGER EXPONENTS

If a is any real number and n is a positive integer then the exponential notation na , readas the nth power of a , is the product of n-factors of a. That is,

a n a a a a an factors

In this notation, a is called the base and n is the exponent . The nth power of some numbers aregiven below.

a) 243)3)(3)(3)(3(33 5

b)16

1

2

1

2

1

2

1

2

1

2

1 4

c)125

8 5

2

5

2

5

2

5

2

3

There will be instances where we encounter an exponential notation of the form na

where n is a negative integer or zero. Provided 0 a and n is a positive integer, then

1 0 a andn

n

aa1

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Examples:

a) 13 0 b) 1)2( 0 c) 12

1

0

d)3

3

2

12 e)

5

5

2

1

1

2

1

To get our way around algebraic expressions involving exponential notations, we should be familiar with the rules that govern them.

Laws of Exponents

For any a and b , real numbers not equal to zero and for any n and m ,positive integers

1. Product of two powers of the same basenmnm aaa

2. Quotient of two powers of the same base

mn

nm

n

m

a

a

a

a

1

1

mnif

nmif

nmif

3. Power of a Power

mnnm aa

4. Power of a Product

mmm baab

5.

Power of a Quotient

m

mm

b

a

b

a

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Example: Simplify each of the following in a form free of zero or negative exponents.

a) 52323 222 y y y y

b) 1

5

20

x

c) 6623323122

01

32

y x y x y x y x

y x

d)22

22

2324

322

23

32

24

2 11

mr rm sm sr

r m

sm

r

sr

m

Exercises: Simplify the following expressions. Assume no denominator is equal to zero.

Express the results with positive exponents.

1.

sr sr 6

45

6. 2

34

6425

93

cbacba

2. 2332 33 x x 7. 27ab3c7

d 2

1

9a3b3c6

d

2

3.

0

33

22

4

8

nm

nm 8.

64 x2 y6 z 6

v3w2

216 x3 y 3 z 6

v2w

3

2 x5 y3 z 6

vw

4. 22

23

)(

)(

x

x

b

b

9.

21 x 2 y3 z 2

vw2

1

6 xyz 3

v2w5

02 x 0 y3 z 4

v3w2

2

5.

1

41

321

6

3

xy

y x

2.4 OPERATIONS WITH ALGEBRAIC EXPRESSIONS

The four fundamental operations ,as well as determining powers and roots, can be carried

out with algebraic expressions.

Addition/Subtraction of Algebraic Expressions:

When two or more algebraic expressions are added, the sum is obtained by adding similar

terms together. The sum of similar terms is the sum of the numerical coefficients multiplied by

the common literal coefficients. To simplify the procedure, the terms may be arranged incolumns so that similar terms lie in one column.

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Likewise, when the difference between any two algebraic expressions is required, the

difference is obtained by subtracting similar terms. The difference of similar terms is the

difference of the numerical coefficients multiplied by the common literal coefficients. Again,similar terms may be arranged in one column for easy manipulation. One may look at subtraction

in terms of addition as equal to the sum of the minuend and the negative of the subtrahend. That

is, A – B = A + (-B).

Examples: Find the indicated sum or difference:

1. Find the sum of the following expressions:

a. xy x y y xy x x y xy 95611,3734,9843 Solution:

3xy – 4y – 8x + 9

3xy - 7y + 4x – 3

-9xy + 6y – 5x +11

-3xy - 5y – 9x + 17

b. y x x xy y xy y x x y x xy y x xy 222222 3867y,373,6494128 Solution:

8 xy2 12 x 2 y 4 xy 9 x 4 y 6

- 7 xy2 3 x 2 y x 3 y

6xy2 3 x 2 y 8x 7y

7 xy2

12 x2

y 4 xy 18 x 6

2. From the sum of 265323 43 y y x xy xy and 12y4 43 y x xy , subtract

7542 43 y xy y x xy .

Solution: Add the first two expressions

265323 43 y y x xy xy

14y2 43 y x xy

259y4 3 4 y x xy ( This is the minuend for subtraction)-

7 5 2 4 43 y y x xy xy ( Subtrahend)

32 14 6 - 43 y y x xy xy

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Exercises:

I. Find the sum of the given expressions.

a. 2 x 3 y 6 z , 5 y 4 z 3 x, z 3 x 5 y

nmqp pqnm pq

mnqp pqmn pq

mnqp pqnm pqmnb

342

324

342

73mn

2nm-

753 .

c. 3rs 4rs2 3r 2 s 12r 2 s2 5rs 3rs2 r 2 s 8r 2 s2 rs rs2 2r 2 s r 2 s2

d. m3n3 2m2n4 4mn5 6n6 5m2n4 8n6 4m3n3 5mn5 3m2n4 2m3n3 3n6 mn5

II. Subtract from the sum of the first two expressions the third expression.

a. xyz 4 x 2 yz 3 xy2 z 4 xyz 2 6 xyz 2 5 xyz 5 xy2 z 8 x 2 yz 2 xy 2 z 3 xyz 2 2 xyz 10 x 2 yz

b. 2abc2ad 4bc ac bc 2ac 5ab 3ab 5ac 5bc abc

III. What must be added to a5b 3ab2 a3 3b4 a 5 to give27 3a5b 5ab2 4a 3a3 2b4 ?

IV. From what polynomial should alm3 5a2lm2 4al 3m 2a 6l 4m be subtractedfrom to give 3m l 3a 2alm3 5a2lm2 2al 3m ?

V. What must be subtracted from the sum of m3in 3mi2n2 4m 6i3n 2i and6m 3m3in 2i mi2n2 6i3n to give 5i 9m 4m3in 6i3n 9mi2n2 ?

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Simplifying Algebraic Expressions by Removal of Symbols of Groupings

There will be occasions where complicated forms of algebraic expressions need to besimplified. Forms of algebraic expressions become complicated when different groupings of

terms are given, some of which maybe embedded within groups. To simplify expressions of this

nature, we perform the indicated operations starting with the innermost group moving out.Similar terms within groups need to be added together before another grouping symbol isremoved.

Examples: Simplify the given expression by removing symbols of grouping:

a. 3 x2 3 xy 2 y2 2 7 y2 x2 xy = x 2 9 xy 6 y 2 14 y 2 2 x 2 2 xy = x 2 7 xy 20 y 2

b. 2 x 9 4 3 x 2 6 x 5 x 2 = 2 x 18 4 3 x 12 2 x 5 x 2 = 2 x 18 4 4 x 10 = 2 x 18 16 x 40 = 14 x 58

c. 3 x 2 x 3 x 1 2 x 4 3 x 2 2 x = 3 x 2 x 3 x 32 x 4 3 x 4 2 x

= 3 x 2 2 x 3 4 x 4 = 3 x 4 x 6 4 x 16 = 3 x 22 = 3 x 66

Exercises: Simplify the given expressions by removing symbols of grouping:

1. 2 x 5 y 7 y 7 x 2. 3 ab 3cd 4 2ba 5cd 3 ab 7cd 3. 3 2w 3 s w s w s 4. 2 x 2 1 3 x 2 x 2 3 2 x 4 x 1

5. a b c 2 ab 3 c d ab 3 z 4 c c 32a 6. 3 2 x 4 3 x 2 x 3 4 3 x

7. 2 x y 2 3 x 4 x 5 y 4 3 y 5 x 3 7 y x 8 2 x y 2 8. a 4 b 6c 2 3 5c 3a 7 b c 2 3b c 4 b c 6c a

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Multiplication of Algebraic Expressions

The product of algebraic expressions is the result of applying the distributive propertyand the laws of exponents, where each and every term of the first factor is multiplied with all

the terms of the second factor. The result is simplified by adding similar terms together.

Examples:

1. Multiply 5)-3yx()4352( 22 y x x xy Solution:

5)-3yx()4352( 22 y x x xy

= x)4352( 22 y x x xy + 3y)4352( 22 y x x xy + 5)-()4352( 22 y x x xy

= x)4352( 2322 y x x y x + )129156( 322 y y x xy xy + )20152510( 22 y x x xy

= 20151295621352 223222322 y x y y x xy xy x y x x y x

2. Multiply y)x()6433( 54322345 y xy y x y x y x x Solution:

y)x()6433( 54322345 y xy y x y x y x x

= x)()6433( 54322345 y xy y x y x y x x + (y))6433( 54322345 y xy y x y x y x x

= )6433( 542332456 xy y x y x y x y x x + )6433( 654233245 y xy y x y x y x y x

= 6542332456 6254223 y xy y x y x y x y x x

Or

y)x()6433( 54322345 y xy y x y x y x x

y x

y xy y x y x y x x

6433 54322345

542332456 6433 xy y x y x y x y x x ( product of the first factor with x)654233245 6433 y xy y x y x y x y x ( product of the first factor with y)

6542332456 6254223 y xy y x y x y x y x x ( product of the two factors)

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The product of a polynomial with another polynomial in the same variable can be

simplified if done in tabular form as follows:

1. Arrange both polynomials in descending order of a chosen variable, inserting missingterms with zero coefficient.

2. Use the coefficients of the terms of the first polynomial , written from left to right, as

column headings of the table and the coefficients of the terms of the second polynomial ,written from top to bottom starting with the second row on the right end of the table, asrow headings.

3. A particular entry of the table is obtained multiplying its column and row headings. This

is done for all entries of the table.4. These entries are added diagonally starting from the upper right entry to the lower left

entry. The diagonal sums are the coefficients of the terms of the product whose degree is

the sum of the degrees of the two polynomials multiplied.

Example: Find the product as specified.

1. Product of 2 x5

3 x4

5 x3

x2

7 x 5 and x3

4 x2

5 x 3

Solution:

2 -3 5 1 -7 5

2 -3 5 1 -7 5 1

8 -12 20 4 -28 20 4

-10 15 -25 -5 35 -25 -5

6 -9 15 3 -21 15 3

2 5 -17 42 -37 -13 58 -46 15

The product is 2 x8 5 x7 17 x6 42 x5 37 x4 13 x3 58 x2 46 x 15

2. (3 x5 x4 y 3 x3 y2 x 2 y3 4 xy4 6 y5) ( 3x 2y)

Solution:

3 -1 3 1 4 -6

9 -3 9 3 12 -18 3

6 -2 6 2 8 -12 2

9 3 7 9 14 -10 -12

The product is 9 x

6

3 x5

y 7 x4

y

2

9 x3

y

3

14 x2

y

4

10 xy5

12 y6

Exercises:

I. Multiply the given expressions:

1. 12 xy2 z 2 5 x3 yz 24 xy2z , 3 xy3z4 2. ab2c abc 2bc2 , 2a 5c

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3. x 3 y 3 , x y

4. m5 n5 , m n 5. a2 ab b2 , a b 6. x 4 + 4x3y+6x2y2 +4xy3 + y4 , x+2y

7. 5 x 4y - x3y2 - 2x2y3 +4xy4 - 5y5 , 2x- 5y

8. 3m7 12m5 6m4 5m2 +3m- 7 , 2m5 2m3 3m2 m 1

II. Multiply 3 x 2 y 3 z with the sum of xy2 5 yz 3 x2 yz and 9 xy2 yz 3 5 x2 yz .

III. Find the product of x2 y 2 and x2 y 2 .IV. Find the required quantity:

a. area of a right triangle whose legs are (2x-1) and (x+4).

b. Perimeter of a rectangle with sides (y-z+2) and (3+4z-y).c. Volume of a rectangular prism with sides (m-n+3), (3n+2-7m) and

( 5m-5n+7).

Division of Algebraic Expressions

The quotient of dividing an algebraic expression by another is based on the divisionalgorithm. That is,

)(

)()(

)(

)()()()()(

x D

x R xQ

x D

x N x R x D xQ x N .

The quotient of an algebraic expression with a monomial divisor is the sum of the

quotients obtained from dividing each of the terms of the dividend by the monomial divisor.

That is,

C

B

C

A

C

B A

.

However, when the divisor is not a monomial, the quotient is obtained by a process of

long division as follows:

1. Arrange the terms of the dividend and the divisor in the descending order of the power

of the variable, including terms with zero coefficient. In case the dividend and divisor are

polynomials in two or more variables, the terms should be arranged in the descending order ofthe power of a chosen variable.

2. Divide the first term of the dividend by the first term of the divisor to get the first term

of the quotient.3. Multiply this term of the quotient with each of the terms of the divisor and subtract the

product from the dividend.

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4. Using the difference obtained in step 3 as new dividend, repeat steps 2 and 3 until such

time that the difference obtained from this step is of degree less than the divisor. This expression

is the remainder.

Examples:

1. Divide 22432532 3 by36126 y x y x y x y x .

Solution:

2322

43

22

25

22

32

22

432532

12423

36

3

12

3

6

3

36126 xy x y

y x

y x

y x

y x

y x

y x

y x

y x y x y x

2. Divide 32 by4456 23 x x x x .

Solution:

1

________

32x

4x2

__________

64-

44-

______ __________

96x

123

4456 32

2

2

23

2

23

x x

x x

x

x x

x x x x

The quotient can be checked by verifying whether

44561)32)(123( 232 x x x x x x ?

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2. Divide 13 by1272 234 x x x x x .

Solution:

2-6x

_ __________

13x- x-

1x3x-

____ __________

3 x

22 x

_ __________ __________

26x2x

12

12072 13 x

2

2

23

23

234

2

2342

x x

x x

x

x x

x x x x x

Exercises: Perform the indicated operations:

1. 35 y8 z 3 5 y2z2 2. (24a5b3c2 +36a8 b5c8 - 60a7 b4c5) 12a4 b2c2

3. m2 n2 m- n

4. 12a2 26a 16 2a 1 5. 9 x2 24 x 16 3 x 4 y 5. p3 q3 ( p q) 6. r 8 t 8 (r t ) 7. 3 x7 x6 2 x5 x4 3 x3 x 2 3 x 5 ( x 3) 8. 3a6 5a4b2 8a3b3 5a2b4 b6

(a b)

9. 12m5 15m4n 18m3n2 22m2n3 6mn4 10n5 (2m 3n) 10. 2a6 5a5b 8a4b2 5a3b3 6a2b4 4ab5 8b6 (2a2 5ab b2)

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Synthetic Division

The process of long division of a polynomial P(x) in one variable divided by a binomialof the form ( x – r ) can be simplified using synthetic division doing the following steps.

1.Arrange the terms of the polynomial in the descending order of the power of the

variable, from the leading term to the constant term supplying missing terms with coefficientzero whenever necessary2. Take note of the numerical coefficients of these terms and arrange them in a row.

3. Bring down the first number in the sequence in a new row and multiply this by r.

4. Add the product to the next number in the previous row and placed the sum in the newrow. Multiply this new number with r.

5. Repeat step 4 until the product is added to the last number of the previous row.

6. The first to the second to the last numbers in the new row are the coefficients of the

terms of the quotient, a polynomial of degree one less than the degree of the dividend, arrange inthe descending order of the power of the variable. The last number is the remainder.

Examples:

1. Divide 2 by3366232 234567 x x x x x x x x .

Solution: The coefficients of the terms arranged in the descending order of the power of x are 2,

-3, 1, -2, -1, 6, 4, 3 and r = 2.By synthetic division:

4 2 6 8 14 40 8

2 -3 1 -2 -1 6 -36 -3 2

2 1 3 4 7 20 4 5

Thus, the quotient of dividing 2 x7 3 x6 x 5 2 x4 x3 6 x2 36 x 3 by x 2 is

4207432 23456 x x x x x x remainder 5.

2. Divide 3 by641625522 23468 x x x x x x .

Solution: The coefficients of the terms arranged in the descending order of the power of

x, including missing terms, are 1, 0, -2, 0, 2, -1, 6, 0, 4 and r = -3.By synthetic division:

-3 9 -21 63 -33 24 -24 721 0 -2 0 -52 25 -16 0 -64 -3

1 -3 7 -21 11 -8 8 -24 8

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Thus, x8 2 x6 52 x4 25 x3 16 x2 64 divided by x 3

is 2488112173 234567 x x x x x x x remainder 8.

3. Divide 42 by188120522 345 x x x x .

Solution I: Since the divisor is not of the form (x-r), the dividend and the divisor have to

be transformed by dividing both the dividend and the divisor by 2 to get

x5 26 x4 60 x3 94 divided by x 2.

It is to the expression x5 26 x4 60 x3 94 divided by x 2 that we apply syntheticdivision.

2 -48 24 48 96

1 -26 60 0 0 -94 21 -24 12 24 48 2

The terms of the quotient obtained by dividing 2 x5 52 x4 120 x3 188 by 2 x 4 are with coefficients given by the numbers in the new row, except for the last number

which will give the remainder after multiplying it by 2 (the factor used to divide the

dividend and divisor). Thus, 2 x5 52 x4 120 x3 188 divided by 2 x 4 is

48241224 234 x x x x remainder 4.

Or:

Solution II: Transform the expression by dividing the divisor by the coefficient of its

leading term, 2. Then proceed with synthetic division, dividing

2 by188120522 345 x x x x as follows:

4 -96 48 96 192

2 -52 120 0 0 -188 _ 2

2 -48 24 48 96 4

The terms of the quotient obtained dividing 2 x5 52 x4 120 x3 188 by 2 x 4 arewith coefficients equal to the numbers of the new row divided by two, except the last number

which is the remainder. Thus, 2 x5 52 x4 120 x3 188 divided by 2 x 4 is

48241224234

x x x x with remainder 4.

The process of synthetic division can be extended to divide a polynomial by another

polynomial provided the dividend and the divisor are in terms of the same variable. The general procedure is described as follows.

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Steps for Extended Synthetic Division:

1. Arrange the terms of the dividend and the divisor in the descending order of the power of the variable, inserting missing terms with zero-coefficient.

Let the degree of the dividend be m and of the divisor be n.

2. Form (n+2) number of rows.3. Arrange the coefficients of the terms of the dividend from step 1 in the first row.There will be (m+1) number of columns.

4. If the coefficient of the leading term of the divisor is 1, change the signs of the terms

of the divisor. Then, starting with the coefficient of the second term, write thecoefficients of this expression downward ( forming a column) at the right end of each

rows starting with the second row. Let this be the (m+2)-th column.

5. Bring down the first number of the first row as first number of the (n+2)th row.

Multiply this number with the first number of the (m+2)-th column and write the product in the same row under column 2. Add the numbers of the second column and

write this as entry of the last row.

6.

Using the numbers of the last row ( moving from right to left) and the numbers of the(m+2)-th column ( moving downward) , multiply numbers in corresponding

positions. Write the products under the next column in the same row as the used

numbers of the (m+2)-th column. Starting with the second column upto the nth

column, a new column will have one term more than the previous term.7. Repeat #6 procedure until the (n-1) column from the right is obtained. The (n-1)

column entry of the last row is the coefficient of the first term of the remainder.

8. Starting with the (n-2)th column from the right, the entries in a new column woulddecrease by one term (starting from the top) until the (m+1)th column.

9. The entries of the last row from the first to the nth column from the right are the

coefficients of the terms of the quotient , a polynomial with degree (m-n). The entries

of the last row from the (n-1)-th column from the right are the coefficients of theterms of the remainder, a polynomial of degree ( n-1).

10. If the coefficient of the leading term of the divisor is not 1, then in doing steps 5 and

6, the entries of the last row should be divided first by the coefficient of the leadingterm of the divisor before the products are obtained. This is done for all entries except

for those entries representing the coefficients of the terms of the remainder.

Examples: Use extended synthetic division to obtain the required quotient.

1. Divide x6 - 3x5 +2x4 +x3 +3x2 - 2x +5 by x2 - 2x +3

Solution:

1 -3 2 1 -13 -2 5

2 -2 -6 -4 -16 2

-3 3 9 6 24 -3

1 -1 -3 -2 -8 -12 29

The quotient is x4 - x3 - 3x2 - 2x- 8 remainder -12x +29

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2. Divide 6 x8 5 x7 3 x6 +24x5 -11x4 +22x3 - 2x2 - 4x +4 by 2x3 +3x2 -x+2

Solution:

6 5 3 24 -11 22 -2 -4 4-9 6 -18 3 -3 -9 -3/2

3 -2 6 -1 1 3 ½

-6 4 -12 2 -2 -6 -1

2 6 -4 12 -2 2 6 -8 -3 -23 -2 6 -1 2 3

The quotient is 2-3x-8x-remainder3+x+x-6x+2x-3x 22345 .

Exercises: Divide the first polynomial by the second using synthetic division.

1. 4 x7 3 x6 2 x5 x4 5 x3 3 x2 7 x 2 , x - 3

2. 121011154 234 mmmm , 3m 3. x8 4 x5 2 x 4 , x 4

4. 432

3142840 x x x x , 5 x 5. 213610 23 aaa , 75 a

6. a8 8a7 28a6 56a5 70a456a3 28a2 8a 1 , a 1 7. 10921 324 cccc , 527 2 cc 8.

25 y

9

15 y6

20 y 10 , y3

2 y2

3 y 1

9. m9 1 , m2 +m+1 10. b9 7b8 2b6 5b3 2b 6 , 2b3 2b 1

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2.5 SPECIAL PR0DUCTS

Certain types of products occur so frequently that they deserve our particular attention.Such products are called special products and they should be learned and memorized in order to

save time in multiplication.

Product of Two Binomials

The product of two binomials equals the product of the first terms plus the product of the

outer terms, the product of the inner terms, and the product of the last terms.

(ax + by)(cx + dy) = acx 2 + adxy +bcxy + bdy

2

Combining similar terms, we have

(ax + by)(cx+dy) = acx

2

+ (ad + bc)xy + bdy

2

Example: Find the indicated product.

1) (3x – 5y)(4x + 7y) = 12x2 + xy – 35y

2

The following diagram may be helpful but all multiplications and additions should be

done mentally.

First : (3x)(4x) _______________ 12x2

Outer + Inner : (3x)(7y) + (-5y)(4x) _____ 21xy + (- 20 xy) = xy

Last: (-5y)(7y)______________ -35y2

2. (9x – 7)(2x – 3) = 18x2 – 41x + 21

F: (9x)(2x) _________________ 18x2

O + I : (9x)(-3) + (-7)(2x) _________ -27x - 14x = - 41x

L: (-7)(-3) __________________ 21

3. (2ab2 + 3c)(5ab

2 – 6c) = 10a

2 b

4 + 3ab

2c – 18c

2

Square of a Binomial

The square of the sum ( or difference) of two binomials equals the square of the first

term plus ( or minus) twice the product of the two terms plus the square of the second term.

(a + b) 2 = a

2 + 2ab + b

2

(a – b) 2= a

2 – 2ab + b

2

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Examples : Perform the indicated operation:

1. 222 )7()7)(2(2)2()72( x x x = 49284 2 x x

2. 233222232 )4()4)(3(2)3()43( t t z z t z = 6324 16249 t t z z

3. 222 )2()2)(4)(2()4()2(4 y x y x y x = 22 4416816 y xy x y x

Product of the Sum and Difference of Two Terms

The product of the sum and difference of two binomials equals the square of the first

term minus the square of the second term.

(a + b)(a – b) = a 2 – b

2

Examples: Find the product of each of the following:

1. 22 )2()7()27)(27( y x y x y x = 22 449 y x

2.

22

3

4

4

3

3

4

4

3

3

4

4

3

bababa

=9

16

16

9 22 ba

3. 6565 22 y y y y = )65()65( 22 y y y y = 222 )65()( y y = 366025 24 y y y = 366025 24 y y y

4. 23223232 pnm pnm pnm d bad bad ba

=

pnm

d ba

624

Cube of a Binomial

The cube of a binomial equals the cube of the first term plus (or minus) thrice the productof the first term squared and the second term plus thrice the product of the first term and the

square of the second term plus (or minus) the cube of the second term.

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(a + b) 3 = a

3 + 3a

2 b + 3ab

2 + b

3

(a – b) 3 = a

3 – 3a

2 b + 3ab

2 – b

3

Examples: Find the product of each of the following:

1. 32233

)4()4)(3(3)4()3(3)3(43 y y x y x x y x = 3223 6414410827 y xy y x x

2. 3323232232332 )7()7)(5(3)7()5(3)5(75 nnmnmmnm =

962346 343735525125 nnmnmm

3. 32233 232322 y y x y x x y x bbabaaba = y y x y x x bbabaa 3223 6128

Square of a Trinomial

The square of a trinomial equals the sum of the squares of the three terms plus twice the product of each term and each term that follows it.

(a + b + c) 2 = a

2 + b

2 +c

2 + 2ab + 2ac + 2bc

Examples: Find the required product.

1. cbcabacbacba 342322422342342 2222 = bcacabcba 2412169164

222

2. 224 w z y x = ))((2)2)((2)4)((2416 2222 w x z x y xw z y x ))(2(2))(4(2)2)(4(2 w z w y z y

= zw yw yz xw xz xyw z y x 4816248416 2222 Alternate Solutions:

1. 22 342342 cbacba

= 22 3)3(42242 ccbaba =

222 9241216164 cbcacbaba 2. 224 w z y x = 224 w z y x

= 22 22424 w z w z y x y x = 2222 4481624168 wwz z wy yz xw xz y xy x

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Exercises: Find the indicated product.

1. )43)(32( x x 3. )32()32( 22 x x x x 2. )25)(94( 22 z xy z y x 4. 322 53 y x

5.232

)43( z t 8. 32

3 ba

6.

2

3

2

2

1

27

cab 9. 232 32 z y x

7. 2222 3232 baba 10. 2)23(35)23(4 y x y x

PASCAL’S TRIANGLE

One technique of finding the nth power of a binomial is by means of Pascal’s Triangle

which is shown below. The numbers in each row are the numerical coefficients of the terms of

the expansion of (x + y)n

for n = 0,1,2,3,…n.

n = 0 1 (x + y)0

n = 1 1 1 (x + y)1

n = 2 1 2 1 ( x + y)2

n = 3 1 3 3 1 (x + y)3

n = 4 1 4 6 4 1 (x + y)4

n = 5 1 5 10 10 5 1 (x + y)5

n = 6 1 6 15 20 15 6 1 (x + y)6

n = 7 1 7 21 35 35 21 7 1 (x + y)7

Notes: 1. The terms should be arranged in descending powers of one variable.

2. For (x – y)n , the signs are alternate plus and minus.

Example. Expand (3x – y)5.

543223455 )3(5)3(10)3(10)3(5)3(3 y y x y x y x y x x y x = 54322345 1590270405243 y xy y x y x y x x

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Exercises: Expand each of the following by using Pascal’s triangle.

1. 62 x

2. 432 ba

3.

7

43 d c

2.6 FACTORING

Factoring is the process of finding two or more expressions whose product is the original

expression. The quantities that are multiplied are called factors. When the factors of a number (orexpression) are one and itself, then the number is called a prime number (or prime polynomial). A

number, greater than 1, that is not prime is called a composite number.

To factor a polynomial completely is to express it as a product of two or more prime

polynomials. Thus,

F(x) = F1(x) · F2(x) · F3(x) · · · Fn(x)

where F1(x), F2(x), F3(x), · · · , Fn(x) are prime polynomials.

FACTORIZATION PATTERNS

Common Factors

If each term of a polynomial has the same number (or expression) as a factor, then thisnumber (or expression) is called a common factor. The distributive law in reverse is then applied.

a

ac

a

ac

a

abaad acab

)( d cbaad acab

Examples: Factor the following:

1. )124(2248 223 aaaaaa

2. )95(327153 232232245243324 z y y x z x z xy z xy z y x z y x

3. 5342 )(42)(18)(12 pnm pnm pnm = 332 7)(32)(6 p pnmnm p

Exercises: Factor the following:

1. 862 2 x x

2.234 1263 x x x

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3. 22222 365442 yxaby xcab 4. pt t pt p 486012 342 5. )2(144)2(120)2(24 23 bababa 6. mmm x x x 48213 12

FACTORS OF BINOMIALS

The special products taken up earlier can be of great help when factoring certain binomials

or even polynomials.

Difference of Two Squares:

The difference of the squares of two numbers is equal to the product of the sum and the

difference of the square roots of the two numbers.

a 2 – b

2 = ( a + b)(a - b)

Examples: Factor the following algebraic expressions.

1. 2222 )3()2(94 y x y x = )32)(32( y x y x .

2. )(53)(53)(259 2 y x y x y x = ( )553)(553 y x y x

3. )116)(116(1256 224 y y y = )116)(14)(14( 2 y y y

Sum and Difference of Two Cubes

Let a and b be real numbers, variables or algebraic expressions. Then,

a 3+ b

3= (a + b)(a

2 – ab + b

2 )

a 3 – b 3 = (a

– b)(a 2 + ab + b 2 )

Examples: Find the factors of the following:

1. 3333 )3()2(278 nmnm = 22 )3()3)(2()2()32( nnmmnm = )964)(32( 22 nmnmnm

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2. 32363 )4()3(6427 baba = 22222 )4()4)(3()3()43( bbaaba = )16129)(43( 4222 bababa

3. 3234612 )5()(125 y x y x = 22242424 )5()5)(()()5( y y x x y x = )255)(5( 424824 y y x x y x

4. 333 )2()2(8)2( y x y x

4)2(2)2(2)2( 2 y x y x y x

5. 323266 )()( y x y x

= ))(( 422422 y y x x y x = ))()(( 4224 y y x x y x y x

Note: 4224 y y x x can be factored by adding and subtracting 22 y x . This will be discussed later.

Alternate Solution: 232366 )()( y x y x

= ))(( 3333 y x y x

= ))()()(( 2222 y xy x y x y xy x y x

= ))()()((

2222

y xy x y xy x y x y x

Sum and Difference of Two Odd Powers

If a and b are real numbers or expressions, then

))(( 1342321 nnnnnnn bbababaababa

))(( 1342321 nnnnnnn bbababaababa

Note: For an + b

n, the terms of the second factor have alternate plus and minus signs and for

an - bn , the signs of the terms of the second factor are all positive. Examples:

1. Factor 128x7 + y

7.

7777 )2(128 y x y x = 6542332456 )2()2()2()2()2()2()2( y y x y x y x y x y x x y x

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= )248163264)(2( 6542332456 y xy y x y x y x y x x y x

2. Factor99 wv .

))(( 8762534435267899 wvwwvwvwvwvwvwvvwvwv

Alternate Solution:

))(()()( 633633333399 wwvvwvwvwv

= ))()(( 633622 wwvvwvwvwv

Exercises: Factor each of the following completely.

1. 44 8116 y x 6. 639 827 cba 2. 637 1219 ut t 7. 55 321024 ba

3.q p

nm 82

8.55

243)( z y x 4.

36125 z x 9. 8416 q p

5. nn ba 66 6449 10. 1)(64 3 wu

FACTORING TRINOMIALS

Certain trinomials can easily be factored by recalling some special products.

Perfect Square Trinomial

Let a and b be real numbers, variables or algebraic expressions. The perfect square

trinomials can then be factored as shown below. Note that the first and last terms are perfectsquares.

a 2 + 2ab + b

2 = (a + b)

2

a 2 - 2ab + b

2 = (a - b)

2

Examples: Factor the following perfect square trinomials.

1.222

)32(9124 y x y xy x 2. )256036(256036 223 aaaaaa

= 2)56( aa

3. 16)32(8)32( 2 y x y x = 24)32( y x = 2)432( y x

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Quadratic Trinomial

Factoring a quadratic trinomial ax2 + bx + c is usually done by trial and error. Factors mx +

r and nx + s must be found such that mn = a and rs = c.

ax

2

+ bx +c = (mx + r )(nx +s)

Note that mn = a

rs = c

ms + nr = b

If a = 1, then ax2 + bx + c may be factored as shown below.

x 2 + bx + c = (x + r)(x + s)

where rs = cr + s = b.

Note: If a, b, c are integers, then ax

2

+ bx + c is factorable with integer coefficients if b

2

– 4acis a non-negative perfect square.

Examples: Factor each of the following:

1. )2)(4(822 x x x x 2. 3108

2 x x = (4x + 3)(2x + 1)3.

222 45369 cabcba = 222 54(9 cabcba )= 9(ab + c)(ab – 5c)

4. 10)1(7)1( 222 z z = 5)1(2)1( 22 z z = )4)(1( 22 z z = (z – 1)(z + 1)(z – 2)(z + 2)

Exercises: Factor the following trinomials completely.

.1 49284 2 y y 2.

567 497025 x x x 3. 1816

24 x x

4. 64489

2

2

t

p

t

p

5. 121115 2 y y 6. 224 49730 y y x x 7. 4236 61712 z z y y 8. 32)2(18)2( 24 y x y x 9. 532 36 p p

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10. 45)73(2)73(8 2 aa

FACTORING BY GROUPING

The terms of a polynomial can be grouped in such a way that each group has a common

factor. To factor this type of polynomial, we begin by grouping those terms that have commonfactors and then we use the distributive law to complete the factoring.

Sometimes the terms of the polynomial can be grouped to form a factorable binomial or

trinomial. In this case, we apply the rules for factoring binomials or trinomials after the terms have

been grouped.

Examples: Factor the following completely.

1. 1 baab = )1()( baab = )1()1( bba = )1)(1( ab

2. cbaacaba 2 = )()( 2 cbaacaba = )()( cbacbaa = )1)(( acba

3. yz xz y xy x 22 2 = )()2( 22 y x z y xy x

= )(2 y x z y x

= z y x y x )()( = ))(( z y x y x

4.222 4425 d bd ba = )44(25 222 d bd ba

= 22 )2(25 d ba = )2(5)2(5 d bad ba = )25)(25( d bad ba

OTHER TYPES OF FACTORING:

If a perfect square monomial is added to and subtracted from a trinomial that is not a

perfect square, then the resulting expression can be factored as a difference of two squares.

Examples:

1. Factor 4224 716 y y x x .Solution: Add and subtract 22 y x .

2242224222 )816(716 y x y y x x y y x x

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= 22222 )4( y x y x = xy y x xy y x )4()4( 2222 = xy y x xy y x 2222 44

2. Factor 922 24 x x .

Solution: Add and subtract 16x2.

922 24 x x = 224 1696 x x x = 222 16)3( x x = x x x x 4)3(4)3( 22 = )34)(34( 22 x x x x

3. Factor 8424 984 bbaa .

Solution: Add and subtract 424 ba 4284248424 4)9124(984 babbaabbaa

= 22242 )2()32( abba

= ( )232)(232( 242242 abbaabba

The factor theorem may be used to find the factors of a polynomial. The factor

theorem states that x-c is a factor of a polynomial f(x) if and only if f(c) = 0. In this case,

synthetic division may be used to determine if the trial divisor is a factor or not.

Examples: Factor the following polynomials by synthetic division.

1. 1632 23 x x x

2 3 -6 1 1

2 5 -1

2 5 -1 0

hence, )1( x is a factor of 1632 23 x x x and the other factor is )152( 2 x x

Thus, 1632 23 x x x = )152)(1( 2 x x x

2. 1036 23 x x x

1 -6 3 10 2

2 -8 -10

1 -4 -5 0 55 5

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1 1 0

-1 -1

1 0

Thus, 1036 23 x x x = )1)(5)(2( x x x

Exercises: Factor the following:

1. wxvwuxuv 105126 2. kjhjhk h 2135915 2 3. 222 91249 z yz y x 4. 4224 41125 bbaa 5. 25164 24 z z 6. 4285

234 x x x x 7. 222 818122 d baba

8. 45 24 x x 9. 652 23 x x x 10. 673 x x

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NAME:___________________________________ DATE:_____________

INSTRUCTOR:____________________________ SECTION:_________

Activity No. 2.1

Classification of Algebraic Expressions

I. Classify each of the algebraic expressions as to whether a polynomial, rational or irrational.

1.21 xy

2. 53 32 y yx x

3.3

427 y x

4. xy y x 721

5. y x

y xy x

22 5

II. Classify each of the following expression according to the number of terms present.

1. xy2

2.252

1 y x

y x

3.22

2 y x

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4. 1345 35 x x x

5. z x 321

III. Classify each of the polynomials according to the nature of the numerical coefficients of the

terms.

1. 12542 y x yz zx

2. 323 2 y x

3. y z x4

175.0

2

1

4.2

3 z x

5. xy y x 721

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NAME:___________________________________ DATE:_____________


Activity No.2.2

Evaluation of Algebraic Expressions

Evaluate each of the following expressions at the given values of the variables.

1. BH

2

1 where B = 30 and H = 5

2.22

y x where x = 3 and y = 4

3.2

where))()((2

1 cba sc sb sa s s

and a=5, b=8 and c= 7

4.21

11

R R

where 5 and 2 21 R R

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5. 22 xaa

b where x = 4, a = 12, and b = 3

6. 222 y x xy where x = -1 and y = 3

7. (5x+4)(3x+2) where x = -2

8. y x

x

6

41

where x = -3 and y = 5

9.)4)(3(

12

3

x x

x where

2

1 x

10. 432234 464 babbabaa where a = 4 and b = 3

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NAME:___________________________________ DATE:_____________


Activity No. 2.3


Simplify each of the following expressions by removing symbols of groupings.

1. x y y x 43432

2. ba xba x 2534 3. nmanmbbanbam 32222 4. x y x x x 53253 5. z z y x y x z y x 243325

6. 432334223 nnmmnm

7. cct mcmt mc 42226

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8. 6343431 x x y y x

9. 32423 r s sr r s

10. mmmmmm 213235

NAME:___________________________________ DATE:_____________INSTRUCTOR:____________________________ SECTION:_________

Activity No. 2.4


1. x y y x y x y x 23442

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2. 2133242 x x x

3. t r r t t r rt 2323323

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4. 1894243334 nmnmmnnmmn

NAME:___________________________________ DATE:_____________INSTRUCTOR:____________________________ SECTION:_________

Activity No. 2.5

Integer Exponents

Simplify each of the following expressions to a form free of negative or zero exponents.

1. 32 23 xy y x

2. 654 2 y y y

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3. 34 x

4. 53202 y x

5.

2

4

33

43

23

82

d

ca

d c

ba

6. 23323 2 baab

7.

3

025

232

cba

cba

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8.

2

036

432

y z x

z y x

9. 2230832 3 z ab z ba

10.132

32312

aaa

aaa

z y x

z y x

NAME:___________________________________ DATE:_____________


Activity No. 2.6

Operations with Algebraic Expressions

Perform the indicated operations then simplify the results.

I. For each of the following sets of expressions:

a. z y x z y x z y x 625 , 52 , 343

b. nmmnmmnmnmnmnnm 222222

268mn ,n62 , 253 c. 532 , 4386 , 1523 34234234 x x x x x x x x x x x

find:

1. the sum of the three expressions

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2. the difference between the sum of the first two expressions diminished by the third expression.

11. The difference between the first expression diminished by the sum of the last two

expressions.

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NAME:___________________________________ DATE:_____________


Activity No. 2.7


Perform the indicated product.

1. y x y x 5235

2. z y x z y x 4232

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3. nmnm 233232

4. z y x z z y x z 32543

5. m z y xm z y x 343232

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NAME:___________________________________ DATE:_____________


Activity No. 2.8


1. )4386()1523( 234234 x x x x x x x x

2. y)x5253xy-()5xyyxy23( 432234432234 y x y x x y x

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II. Use method of long division to find the quotient, dividing the first expression by the

second expression.

1. abababba 3 , 36213 32

2. 32243362 5x- , 20155 y y x y x y x

3. 32128966432 2x1 , 4432 y y x y x y x y x

4. 43 , 8-3yy-7y5y-y-253 22345678 y y y y y

III. Use synthetic division to find the quotient dividing the first expression by the second:

y 8 7 by y-2

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NAME:___________________________________ DATE:_____________


Activity No. 2.9

Special Products

Find the indicated product.

1. 753 y y

2. 23 265 baba

3. pqmnq pnm 392 232

4. 225 y x

5. 222 94 ba

6. 2)( z y x

7. 5353 22 x x

8.

a

b

b

a

a

b

b

a

3

2

4

5

3

2

4

5

9. 4343 22 x x x x

10. 334 v z

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11.

3

23

nm

12. 222 w z y x

13. uuuuuuuu 3232

14. 2332 y x y x

15. 2423 q p

16. 1111 22 hh

17. 22 11 bb

18.

21

cc

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NAME:___________________________________ DATE:_____________


Activity No. 2.10

Special Products


1. 4332 x x

2. z y x z xy 2594 22

3. 232 43 t z

4.

2

3

2

2

1

27

cab

5. 2222 3232 baba

6 . 3232 22 x x x x

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7. 322 53 y x

8. 333 bm

9. 232 32 z y x

. 2)23(35)23(4 y x y x

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NAME:___________________________________ DATE:_____________


Activity No. 2.11

Special Products


1. 3)35( r p

2. 234 )3( nm cb

3.

2

325

3

2

pr

4. ))(( y x y x

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5. 1)2(51)2(5 y x y x

6.

68

5

34

2222 nmnm

7. )1)(1( z y x z y x

8. (p2

+ q2 – r

2)(p

2 – q

2+ r

2)

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NAME:___________________________________ DATE:_____________


Activity No. 2.12

Factoring

A. Factor the following by taking out the common factors.

1.7456 742 nmnm

2. 8910 483276 y y y

3. 255 p

4.494385267 3212864 cbacbacba

5. 22422222 )(5)(2 y x y x y x x

6. 21

2

1

4 )1(2)1(

x x x x

7. 31

3

2

2

1

4 xy y x

8.11 44 nn x x

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B. Factor the following binomials completely.

1. 400121 2 p

2.246 4936 u z v

3.22 81225 ba

4. 814 n y

5. 88 y x

6. 273 y

7. 648 6 x

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NAME:___________________________________ DATE:_____________


Activity No. 2.13

Factoring

Factor the following binomials completely.

1. 162 3 z

2. 33 1258 r t

3.1212 nm

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4. 3)(1000 r q

5.15105243 cba

6. 1287 q

7. 333 )()( nmnm

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NAME:___________________________________ DATE:_____________


Activity No. 2.14

Factoring

Express the following trinomials in factored form.

1. 7132 24 z z

2.422244 81818 ccbaba

3.2345 20244 nmnmm

4. 45122 cc

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5. 569180 2 hh

6. 49284 2 aa

7. 25309 24 k k

8.100

81

25

36

25

16 2 y y

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NAME:___________________________________ DATE:_____________


Activity No. 2.15

Factoring

Factor the following expressions completely.

1. 81)2(18)2( 2 y x y x

2. qqnn p pmm 22 44

3. 2022 xy y x

4. 562 aa qq

5. 257049 2 mm

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6.16

14 2 t t

7. 256036 2 z z

8. 22 7176 vuvu

9. 15a4 b

4 + 7a

2 b

2c

2 – 2c

4

10. 14x6y

2 + 41x

3yz + 15z

2

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NAME:___________________________________ DATE:___________


Activity No. 2.16

Factoring

Factor the following by grouping, by adding or subtracting a monomial term or by syntheticdivision.

1.222 496 t srsr

2.222

91249 z yz y x

3. decebecd cbc 161281296 2

4.3322 2 bababa

5.4224 716 d d cc

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6. 25164 24 x x

7. 20918 2 x x

8. 122 23 x x x

9. 653856 234 y y y y

10. 24 )23( x x A. Factor the following by taking out the common factors.

1.7456 742 nmnm

2. 8910 483276 y y y

B. Factor the following binomials completely.

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1. 400121 2 p

2.246 4936 u z v

Factor the following binomials completely.

1. 162 3 z

2.33 1258 r t

Express the following trinomials in factored form.

1. 7132 24 z z

2.

422244

81818 ccbaba

Factor the following expressions completely.

1. 81)2(18)2( 2 y x y x

2. qqnn p pmm 22 44

Factor the following by grouping, by adding or subtracting a monomial term or by synthetic

division.

1. 222 496 t srsr

2. 222 91249 z yz y x

3. decebecd cbc 161281296 2

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4. 3322 2 bababa

5.4224 716 d d cc

6. 25164 24 x x

7. 20918 2 x x

8. 122 23 x x x

9. 653856 234 y y y y

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)23( x x

chapter 2 (math 10-3)

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