# How do you factor completely 12x^2 - 10x - 8?

Apr 7, 2016

$2 \left(3 x - 4\right) \left(2 x + 1\right)$

#### Explanation:

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

Notice that the whole thing is exactly divisible by 2

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

I have already experimented on paper!

Whole number factors of 6$\to \left\{1 , 6\right\} \text{ , } \left\{2 , 3\right\}$

Whole number factors of 4$\to \left\{1 , 4\right\} \text{ , } \left\{2 , 2\right\}$

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

$\left(3 \times 4\right) - \left(2 \times 1\right) = 10 \to$ No good!

$\left(3 \times 1\right) - \left(2 \times 4\right) = - 5 \to$ Got it!

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

$= 2 \left(6 {x}^{2} - 5 x - 4\right)$
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$\textcolor{b l u e}{\text{The factors are: } 2 \left(3 x - 4\right) \left(2 x + 1\right)}$