How do you "factor by grouping" in Pre-Calc?

Question

How do you "factor by grouping" in Pre-Calc?

My pre-calc packet asks me to "factor by grouping".. and I am a bit confused.

Here are some of the questions that tell me to "factor by grouping":
1. x^3 - x^2 + 2x - 2
2. 2x^3 - x^2 - 6x + 3
3. 6x^3 - 2x + 3x^2 - 1

If you could explain it to me it would really help, thanks!

Answer 1

Question 1
x ³ + 2 x - (x ² + 2)
x (x ² + 2) - (x ² + 2)
(x - 1) (x ² + 2)

Question 2
2 x ³ - 6 x - x ² + 3
( 2 x )( x ² - 3 ) - ( x ² - 3 )
( x ² - 3 ) ( 2 x - 1 )

Question 3
( 3 x ² )( 2 x + 1 ) - ( 2 x + 1 )
( 2 x + 1 ) ( 3 x ² - 1 )

Answer 2

The idea of grouping is to make 2 groups (by placing parentheses) of two terms each, in such a way that each group has a common factor. At that point, the common factor can be factored out. Example:
1.
x^3 - x^2 + 2x - 2 =
(x^3 - x^2) + (2x-2) =
x^2(x-1) + 2(x-1) =
(x^2+2)(x-1)

2.
2x^3 - x^2 - 6x + 3=
2x^3 - 6x - x^2 + 3 = (terms rearranged)
(2x^3 - 6x) - (x^2-3)
2x(x^2-3) - 1(x^2-3) =
(2x-1)(x^2-3)

3.
6x^3 - 2x + 3x^2 - 1 =
6x^3 + 3x^2 - 2x - 1 =
(6x^3 + 3x^2) + (-2x - 1)
3x^2(2x+1) - 1(2x+1) =
(3x^2-1)(2x+1)

After some practice you might realize that you end up grouping either the first two terms and the last two terms, or term 1 with 3 and term 2 with 4. This simply makes sure that the two groups when factored have a factor of the same degree, as:

3 - 1 = 2 - 0
3 - 2 = 1 - 0

Answer 3

Factor By Grouping Practice