# How do you factor 6a^2-50ab+16b^2?

Jun 13, 2015

$6 {a}^{2} - 50 a b + 16 {b}^{2} = 2 \left(3 a - b\right) \left(a - 8 b\right)$

#### Explanation:

$6 {a}^{2} - 50 a b + 16 {b}^{2}$

$= 2 \left(3 {a}^{2} - 25 a b + 8 {b}^{2}\right)$

Use a version of AC Method:

Let $A = 3$, $B = 25$, $C = 8$

Look for a factorization of $A C = 3 \cdot 8 = 24$ into two factors whose sum is $B = 25$. The pair $B 1 = 1$, $B 2 = 24$ 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 get the coefficients of a factor of the quadratic:

$\left(A , B 1\right) = \left(3 , 1\right) \to \left(3 , 1\right) \to \left(3 a - b\right)$
$\left(A , B 2\right) = \left(3 , 24\right) \to \left(1 , 8\right) \to \left(a - 8 b\right)$

Hence:

$3 {a}^{2} - 25 a b + 8 {b}^{2} = \left(3 a - b\right) \left(a - 8 b\right)$

and

$6 {a}^{2} - 50 a b + 16 {b}^{2} = 2 \left(3 a - b\right) \left(a - 8 b\right)$