# How do you factor 2x^3 - 3x^2 - 5x?

Nov 14, 2015

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

#### Explanation:

First, note that each term has a factor of $x$, and so we have
$2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 {x}^{2} - 3 x - 5\right)$

Now, we can use the quadratic formula to find the remaining factors, but first let's see if there are easy integer solutions by looking for $a , b , c , d$ where $2 {x}^{2} - 3 x - 5 = \left(a x + b\right) \left(c x + d\right)$

We know that $a c = 2$ and so we can look at $\left(2 x + b\right) \left(x + d\right)$

We also know $b d = - 5$ and so our possible choices are $\left(2 x + 1\right) \left(x - 5\right)$ and $\left(2 x - 5\right) \left(x + 1\right)$

Multiplying these out shows that $2 {x}^{2} - 3 x - 5 = \left(2 x - 5\right) \left(x + 1\right)$

So our final result is
$2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 x - 5\right) \left(x + 1\right)$