How do you factor  6y^2 + 27y - 15?

Sep 19, 2016

$6 {y}^{2} + 27 y - 15 = 3 \left(2 y - 1\right) \left(y + 5\right)$

Explanation:

First note that all of the coefficients are divisible by $3$, so we can separate that out as a factor:

$6 {y}^{2} + 27 y - 15 = 3 \left(2 {y}^{2} + 9 y - 5\right)$

We can factor the quadratic $2 {y}^{2} + 9 y - 5$ using an AC method:

Find a pair of factors of $A C = 2 \cdot 5 = 10$ which differ by $B = 9$

The pair $10 , 1$ works.

Use this pair to split the middle term and factor by grouping:

$2 {y}^{2} + 9 y - 5 = \left(2 {y}^{2} + 10 y\right) - \left(y + 5\right)$

$\textcolor{w h i t e}{2 {y}^{2} + 9 y - 5} = 2 y \left(y + 5\right) - 1 \left(y + 5\right)$

$\textcolor{w h i t e}{2 {y}^{2} + 9 y - 5} = \left(2 y - 1\right) \left(y + 5\right)$

Putting it together:

$6 {y}^{2} + 27 y - 15 = 3 \left(2 y - 1\right) \left(y + 5\right)$