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How do you factor $6 x^{3} + 34 x^{2} - 12 x$ $6x^3+34x^2-12x$ ?

1 Answer

Oct 9, 2015

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

Explanation:

Given
$XXX 6 x^{3} + 34 x^{2} - 12 x$ $color(white)("XXX")6x^3+34x^2-12x$

First extract the obvious common factor of $(2 x)$ $(2x)$ from each term:
$XXX (2 x) (3 x^{2} + 17 x - 6)$ $color(white)("XXX")(2x)(3x^2+17x-6)$

In the hopes of finding integer coefficient factors of $(3 x^{2} + 17 x - 6)$ $(3x^2+17x-6)$
we consider the integer factors of $3$ $3$ and of $(- 6)$ $(-6)$
looking for a combination that will give us a sum of products $= 17$ $=17$

There are only a few possibilities. The diagram below might help understand the process:
enter image source here
The "working" combination is $(3 x - 1) (1 x + 6)$ $(3x-1)(1x+6)$

Which gives us the complete factorization:
$XXX 6 x^{3} + 34 x^{2} - 12 x$ $color(white)("XXX")6x^3+34x^2-12x$
$XXXXXXXXXXX = (2 x) (3 x - 1) (1 x + 6)$ $color(white)("XXXXXXXXXXX")=(2x)(3x-1)(1x+6)$

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