Chapter 6 FaCtoring and the distributive ProPerty 147
- 2(x^2 − 6)^9 + (x^2 − 6)^4 + 4(x^2 − 6)^3 + (x^2 − 6)^2
= 2 (x^2 − 6 )^7 (x^2 − 6)^2 + (x^2 − 6 )^2 (x^2 − 6)^2 + 4 (x^2 − 6 )(x^2 − 6)^2 + 1 (x^2 − 6)^2
= [2(x^2 − 6 )^7 + (x^2 − 6 )^2 + 4 (x^2 − 6 ) + 1 ](x^2 − 6)^2
= [2(x^2 − 6)^7 + (x^2 − 6)^2 + 4x^2 − 24 + 1](x^2 − 6)^2
= [2(x^2 − 6)^7 + (x^2 − 6)^2 + 4x^2 − 23](x^2 − 6)^2
- (15xy − 1)(2x − 1)^3 − 8(2x − 1)^2 = ( 15 xy − 1 )( 2 x − 1 )(2x − 1)^2 − 8 (2x − 1)^2
= [( 15 xy − 1 )( 2 x − 1 ) − 8 ](2x − 1)^2
Factoring by Grouping
Some expressions having four terms can be factored with a technique called
factoring by grouping. Factoring by grouping works to factor an expression con-
taining four terms if we can factor the first two terms and the last two terms so
that these pairs have a common factor. We begin by factoring the first two
terms; next we see if we can factor the last two terms in a similar way.
EXAMPLES
Use factoring by grouping to factor the expression.
3 x^2 – 3 + x^3 – x
Let us begin by factoring 3 from the first two terms.
33 xx^23 –+ –=xx 31 ()^23 –+xx–
We now see if we can factor the last two terms so that x^2 – 1 is a common
factor. We will factor x from each of the last two terms.
33 31
31 1
23 23
22
xxxx xx
xxx
–+ –= –+–
=–+–
()
()()
Now we can factor x^2 – 1 from each term.
33 31
31 1
1
23 23
22
2
xxxx xx
xxx
x
–+ –= –+–
=–+–
=–
()
()()
()(33+x)
3 xy – 2y + 3x^2 – 2x = y(3x – 2) + x(3x – 2) = (y + x)(3x – 2)
5 x^2 – 25 – x^2 y + 5y = 5(x^2 – 5) – y(x^2 – 5) = (5 – y)(x^2 – 5)
4 x^4 + x^3 – 4x – 1 = x^3 (4x + 1) – (4x + 1) = (x^3 – 1)(4x + 1)
EXAMPLES
Use factoring by grouping to factor the expression.
