# How do you factor 7x^2-10x-8?

Jun 3, 2015

I will use a version of the AC Method...

$A = 7$, $B = 10$, $C = 8$

Look for a pair of factors of $A C = 56$ whose difference is $B = 10$

The pair $B 1 = 4$, $B 2 = 14$ works.

Then for each of the pairs $\left(A , B 1\right)$ and $\left(A , B 2\right)$ divide by the HCF (highest common factor) to find the coefficients of a factor of the quadratic...

$\left(A , B 1\right) = \left(7 , 4\right) \to \left(7 , 4\right) \to \left(7 x + 4\right)$
$\left(A , B 2\right) = \left(7 , 14\right) \to \left(1 , 2\right) \to \left(x - 2\right)$

So $7 {x}^{2} - 10 x - 8 = \left(7 x + 4\right) \left(x - 2\right)$