# How do you factor completely 2y^3 + 6y^2 - 5y - 15?

Aug 1, 2018

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

#### Explanation:

There are $4$ terms. Make two groups which each have a common factor.

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

$= 2 {y}^{2} \left(y + 3\right) - 5 \left(y + 3\right) \text{ } \leftarrow$ there is a common bracket

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

Aug 1, 2018

$\left(y + 3\right) \left(\sqrt{2} y - \sqrt{5}\right) \left(\sqrt{2} y + \sqrt{5}\right)$

#### Explanation:

$\text{factor first/second and third/fourth terms by "color(blue)"grouping}$

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

$\text{take out the "color(blue)"common factor } \left(y + 3\right)$

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

$\text{factor "(2y^2-5)" as a "color(blue)"difference of squares}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$\text{with "a=sqrt2y" and } b = \sqrt{5}$

$= \left(y + 3\right) \left(\sqrt{2} y - \sqrt{5}\right) \left(\sqrt{2} y + \sqrt{5}\right)$