# How do you solve 2x^3 – 5x^2 – 3x + 2 = 2?

Jan 21, 2016

$x = - \frac{1}{2} , 0 , 3$

#### Explanation:

Begin by subtracting $2$ from both sides of the equation.

$2 {x}^{3} - 5 {x}^{2} - 3 x = 0$

Factor an $x$ from the terms on the left hand side.

$x \left(2 {x}^{2} - 5 x - 3\right) = 0$

$x \left(2 x + 1\right) \left(x - 3\right) = 0$
Now that you have these three terms being equal to $0$, you know that only one of them has to equal $0$ for the entire expression to equal $0$. Thus, the three solutions can be found by setting each term equal to $0$:
$x = 0$
$2 x + 1 = 0 \text{ "=>" } x = - \frac{1}{2}$
$x - 3 = 0 \text{ "=>" } x = 3$