How do you factor completely $12 x^{2} - 10 x - 8$ $12x^2 - 10x - 8$ ?

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How do you factor completely $12 x^{2} - 10 x - 8$ $12x^2 - 10x - 8$ ?

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Answer 1 · 2016-04-07T20:03:35

$2 (3 x - 4) (2 x + 1)$ $2(3x-4)(2x+1)$

Explanation:

Some times it is a matter of looking for common factors and experimenting.

Notice that the whole thing is exactly divisible by 2

$2 (6 x^{2} - 5 x - 4)$ $2(6x^2-5x-4)$

I have already experimented on paper!

Whole number factors of 6 $\to {1, 6}, {2, 3}$ $-> {1,6}" , "{2,3}$

Whole number factors of 4 $\to {1, 4}, {2, 2}$ $-> {1,4}" , "{2,2}$

Of these you are looking for a combination that will sum to $- 5 x$ $-5x$

$(3 \times 4) - (2 \times 1) = 10 \to$ $(3xx4)-(2xx1) = 10->$ No good!

$(3 \times 1) - (2 \times 4) = - 5 \to$ $(3xx1)-(2xx4)=-5->$ Got it!

$2 (3 x - 4) (2 x + 1) = 2 (6 x^{2} + 3 x - 8 x - 4)$ $2(3x-4)(2x+1)" "=" " 2(6x^2+3x-8x-4)$

$= 2 (6 x^{2} - 5 x - 4)$ $=2(6x^2-5x-4)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$The factors are: 2 (3 x - 4) (2 x + 1)$ $color(blue)("The factors are: "2(3x-4)(2x+1))$

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