# How do you factor completely: 18x^2 − 21x − 15?

Jul 13, 2015

$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right) = 3 \left(2 x + 1\right) \left(3 x - 5\right)$

#### Explanation:

First notice the common scalar factor $3$ and separate that out:

$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right)$

Then factor $6 {x}^{2} - 7 x - 5$ using a variant of the AC Method.

Let $A = 6$, $B = 7$, $C = 5$.

Look for a pair of factors of $A C = 30$ whose difference is $B = 7$.

The pair $B 1 = 10$, $B 2 = 3$ works.

Next for each of the pairs $\left(A , B 1\right)$, $\left(A , B 2\right)$ divide by the HCF (highest common factor) to give a pair of coefficients of a linear factor, choosing suitable signs:

$\left(A , B 1\right) = \left(6 , 10\right) \to \left(3 , 5\right) \to \left(3 x - 5\right)$
$\left(A , B 2\right) = \left(6 , 3\right) \to \left(2 , 1\right) \to \left(2 x + 1\right)$

Hence:

$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right) = 3 \left(2 x + 1\right) \left(3 x - 5\right)$