# Is it possible to factor f(y)=2y^3+7y^2-30y ? If so, what are the factors?

Nov 8, 2017

$f \left(y\right) = y \left(2 y - 5\right) \left(y + 6\right)$

#### Explanation:

Given:

$f \left(y\right) = 2 {y}^{3} + 7 {y}^{2} - 30 y$

Note that all of the terms are divisible by $y$. So we can separate that out as a factor:

$2 {y}^{3} + 7 {y}^{2} - 30 y = y \left(2 {y}^{2} + 7 y - 30\right)$

Next try an AC method to factor the remaining quadratic:

Find a pair of factors of $A C = 2 \cdot 30 = 60$ which differ by $B = 7$

(We look for difference rather than sum since the coefficient of the constant term is negative)

The pair $12 , 5$ works.

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

$2 {y}^{2} + 7 y - 30 = \left(2 {y}^{2} + 12 y\right) - \left(5 y + 30\right)$

$\textcolor{w h i t e}{2 {y}^{2} + 7 y - 30} = 2 y \left(y + 6\right) - 5 \left(y + 6\right)$

$\textcolor{w h i t e}{2 {y}^{2} + 7 y - 30} = \left(2 y - 5\right) \left(y + 6\right)$

So:

$f \left(y\right) = y \left(2 y - 5\right) \left(y + 6\right)$