# How do you factor 11m^2 + 14m -16?

Jun 4, 2015

I will use a version of the AC Method.

Given $f \left(m\right) = 11 {m}^{2} + 14 m - 16$

$A = 11$, $B = 14$, $C = 16$

Look for a factorization of $A C = 11 \times 16 = 176$ into a pair of factors whose difference is $B = 14$.

The pair $B 1 = 22$, $B 2 = 8$ 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 derive the coefficients of a factor of $f \left(m\right)$, choosing signs appropriately...

$\left(A , B 1\right) = \left(11 , 22\right) \to \left(1 , 2\right) \to \left(m + 2\right)$
$\left(A , B 2\right) = \left(11 , 8\right) \to \left(11 , 8\right) \to \left(11 m - 8\right)$

Hence:

$11 {m}^{2} + 14 m - 16 = \left(m + 2\right) \left(11 m - 8\right)$