How do you factor by grouping four-term polynomials and trinomials?

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Answer 1 · 2014-11-30T04:55:40

Factoring by grouping involves grouping terms then factoring out common factors. Here are examples of how to factor by grouping:

Example with trinomial:
$3 x^{2} - 16 x - 12$ $3x^2 - 16x - 12$ , where $a x^{2} = 3 x^{2}, b x = - 16 x, c = - 12$ $ax^2 = 3x^2, bx = -16x, c=-12$ .

To use grouping method you need to multiply $a x^{2}$ $ax^2$ and $c$ $c$ , which is $- 36 x^{2}$ $-36x^2$ in this example. Now you need to find two terns that multiplied gives you $- 36 x^{2}$ $-36x^2$ but add to -16x. Those terms are -18x and 2x. We now can replace $b x$ $bx$ with those two terms:
$3 x^{2} - 16 x - 12$ $3x^2 - 16x - 12$
$3 x^{2} - 18 x + 2 x - 12$ $3x^2 - 18x + 2x - 12$
Group the expression by two:
$(3 x^{2} - 18 x) + (2 x - 12)$ $(3x^2 - 18x) + (2x - 12)$
Factor out GCF in each group:
$3 x (x - 6) + 2 (x - 6)$ $3x(x - 6) + 2(x - 6)$
(The binomials in parentheses should be the same, if not the same... there is an error in the factoring or the expression can not be factored.)
The next step is factoring out the GCF which basically has you rewrite what is in parentheses and place other terms left together:
$(x - 6) (3 x + 2)$ $(x - 6)(3x +2)$ (THE ANSWER)

Example with polynomial:
$x y - 3 x - 6 y + 18$ $xy - 3x - 6y + 18$

Group the expression by two:
$(x y - 3 x) - (6 y - 18)$ $(xy - 3x) - (6y - 18)$
Careful with the sign outside before parenthesis.. changes sign of the 18.
Factor out GCF in each group:
$x (y - 3) - 6 (y - 3)$ $x(y - 3) - 6(y - 3)$
(The binomials in parentheses should be the same, if not the same... there is an error in the factoring or the expression can not be factored.)
The next step is factoring out the GCF which basically has you rewrite what is in parentheses and place other terms left together:
(y - 3)(x - 6) (THE ANSWER)