# How do you factor completely 20y^2 + 3y - 2?

Jan 9, 2017

$20 {y}^{2} + 3 y - 2 = \left(4 y - 1\right) \left(5 y + 2\right)$

#### Explanation:

Given:

$20 {y}^{2} + 3 y - 2$

Use an AC method:

Find a pair of factors of $A C = 20 \cdot 2 = 40$ which differ by $B = 3$

The pair $8 , 5$ works.

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

$20 {y}^{2} + 3 y - 2 = \left(20 {y}^{2} + 8 y\right) - \left(5 y + 2\right)$

$\textcolor{w h i t e}{20 {y}^{2} + 3 y - 2} = 4 y \left(5 y + 2\right) - 1 \left(5 y + 2\right)$

$\textcolor{w h i t e}{20 {y}^{2} + 3 y - 2} = \left(4 y - 1\right) \left(5 y + 2\right)$