# How do you factor 6x^3+21x^2+15x ?

Jun 26, 2015

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

$= 3 x \left(2 x + 5\right) \left(x + 1\right)$

#### Explanation:

Given $6 {x}^{3} + 21 {x}^{2} + 15 x$

First notice that all of the terms are divisible by $3 x$, so separate out that as a factor first:

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

Let $f \left(x\right) = 2 {x}^{2} + 7 x + 5$

First notice that $2 - 7 + 5 = 0$, so $f \left(- 1\right) = 0$, so $\left(x + 1\right)$ is a factor of $f \left(x\right)$

The remaining factor must be $\left(2 x + 5\right)$ in order that the coefficient $2$ of ${x}^{2}$ and the constant term $5$ result:

$2 {x}^{2} + 7 x + 5 = \left(2 x + 5\right) \left(x + 1\right)$

Putting it all together:

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