copyright © 2007 pearson education, inc. slide r-1


Chapter R: Reference: Basic Algebraic Concepts

R.1 Review of Exponents and Polynomials

R.2 Review of Factoring

R.3 Review of Rational Expressions

R.4 Review of Negative and Rational Exponents

R.5 Review of Radicals


R.2 Review of Factoring

• Factoring is the process of finding polynomials, called factors, whose product equals a given polynomial.

• A polynomial that cannot be written as a product of polynomials with integer coefficients is a prime or irreducible polynomial.

• A polynomial is factored completely when it is written as a product of prime polynomials with integer coefficients.


R.2 Factoring Out the Greatest Common Factor

Example Factor out the greatest common factor.

(a) (b)

Solution

(a)

(b)

5 29y y 26 8 12x t xt t

5 2 2 3 2 2 39 (9 ) (1) (9 1)y y y y y y y

2 26 8 12 2 (3 4 6)x t xt t t x x


R.2 Factoring by Grouping

Example Factor each polynomial by grouping.

(a) (b)

Solution

(a)

2 27 3 21mp m p 3 24 2 2 1x x x

2 2 2 2

2 2

2

7 3 21 ( 7 ) (3 21)

( 7) 3( 7)

( 7)( 3)

mp m p mp m p

m p p

p m


R.2 Factoring by Grouping

Solution

(b) 3 2 2

2

4 2 2 1 2 (2 1) 1(2 1)

(2 1)(2 1)

x x x x x x

x x


R.2 Factoring Trinomials

Example Factor each trinomial.

(a)

(b)

24 11 6y y

26 7 5p p



Solution (a) We must find integers a, b, c, d such that

By FOIL, we see that ac = 4 and bd = 6. Thus a and c are 1 and 4 or 2 and 2.

Since the middle term is negative, consider only negative values for b and d. The possibilities are –1 and –6 or –2 and –3.

24 11 6 ( )( )y y ay b cy d



Solution (a) Try these combinations of factors

The last trial gives the correct factorization.

2

2

2

(2 1)(2 6) 4 14 6

(2 2)(2 3) 4 10 6

( 2)(4 3) 4 11 6

y y y y

y y y y

y y y y



Solution (b) Try the various combinations of factors

The last trial gives the correct factorization.

2

2

2

(2 5)(3 1) 6 13 5

(2 5)(3 1) 6 13 5

(3 5)(2 1) 6 7 5

p p p p

p p p p

p p p p


R.2 Factoring Special Products

Perfect Square Trinomials

2 2 2

2 2 2

2 ( )

2 ( )

x xy y x y

x xy y x y


R.2 Factoring Perfect Square Trinomials

Example Factor each polynomial.

(a) (b)

Solution

(a)

(b)

2 216 40 25p pq q

2 2 4 2 2 2 2

2 2

169 104 16 (13 ) 2(13 )(4 ) (4 )

(13 4 )

x xy y x x y y

x y

2 2 2 2

2

16 40 25 (4 ) 2(4 )(5 ) (5 )

(4 5 )

p pq q p p q q

p q

2 2 4169 104 16x xy y



Difference of Squares

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


R.2 Factoring Differences of Squares


(a) (b)

Solution

(a)

(b)

24 9m 2 46 9x x y

2 2 24 9 (2 ) 3 (2 3)(2 3)m m m m

2 4 2 2 2

2 2

2 2

6 9 ( 3) ( )

[( 3) ][( 3) ]

( 3 )( 3 )

x x y x y

x y x y

x y x y



Difference and Sums of Cubes

Difference of cubes

Sum of cubes

3 3 2 2( )( )x y x y x xy y

3 3 2 2( )( )x y x y x xy y


R.2 Factoring Sums and Differences of Cubes


(a) (b)

Solution

(a)

3 27x 3 364m n

3 3 3

2 2

2

27 3

( 3)( 3 3 )

( 3)( 3 9)

x x

x x x

x x x


R.2 Factoring Sums and Differences of Cubes

Solution

(b)3 3 3 3

2 2

2 2

64 (4 )

( 4 )[ (4 ) (4 ) ]

( 4 )( 4 16 )

m n m n

m n m m n n

m n m mn n


R.2 Factoring by Substitution

Example Factor the polynomial

Solution Replacing 2a – 1 with m and factoring gives

Now, replace m with 2a – 1 in the factored form and simplify.

210(2 1) 19(2 1) 15 .a a

210 19 15 (5 3)(2 5)m m m m


R.2 Factoring by Substitution

Solution

210(2 1) 19(2 1) 15

[5(2 1) 3][2(2 1) 5]

(10 5 3)(4 2 5)

(10 2)(4 7)

2(5 1)(4 7) .

a a

a a

a a

a a

a a

copyright © 2007 pearson education, inc. slide r-1

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