# How do you factor 10x^2 - 7x - 12?

May 31, 2015

Use a variant of the AC Method.

$A = 10$, $B = 7$, $C = 12$

Look for a pair of factors of $A C = 10 \times 12 = 120$ whose difference is $7$. We look for the difference rather than the sum because the sign of the constant term is negative.

The pair $15 , 8$ works:

$15 \times 8 = 120$
$15 - 8 = 7$

Now use that pair to split the middle term, then factor by grouping...

$10 {x}^{2} - 7 x - 12 = 10 {x}^{2} - 15 x + 8 x - 12$

$= \left(10 {x}^{2} - 15 x\right) + \left(8 x - 12\right)$

$= 5 x \left(2 x - 3\right) + 4 \left(2 x - 3\right)$

$= \left(5 x + 4\right) \left(2 x - 3\right)$

May 31, 2015

The method George used is the popular factoring AC Method (YouTube). To avoid the lengthy factoring by grouping, you may use the new AC Method.

$f \left(x\right) = 10 {x}^{2} - 7 x - 12 = 10 \left(x + p\right) \left(x + q\right) .$
Convert y to $y ' = {x}^{2} - 7 x - 120$ = (x + p')(x + q')
Find p' and q' by composing factor pairs of (a.c = -120). a and c have different signs. Proceed: ...(-4, 30)(-5, 24)(-8, 15). This sum is 7 = -b. Then p' = 8 and q' = -15.
Then, $p = \frac{p '}{a} = \frac{8}{10} = \frac{4}{5}$ and $q = - \frac{15}{10} = - \frac{3}{2.}$

Factored form: $f \left(x\right) = 10 \left(x + \frac{4}{5}\right) \left(x - \frac{3}{2}\right) = \left(5 x + 4\right) \left(2 x - 3\right) .$

Check by developing: f(x) = 10x^2 - 15x + 8x - 12. OK.

This new AC Method is fast, systematic, no guessing, no factoring by grouping.