# What is the solution set for -x^3-2x^2=-x-2?

Aug 22, 2015

The solution set is $x = \left\{- 2 , - 1 , 1\right\}$

#### Explanation:

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

Multiply the equation times $- 1$.

${x}^{3} + 2 {x}^{2} = x + 2$

Move all terms to the left side by subtracting $x + 2$ from both sides.

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

Group the terms into two binomials.

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

Factor out ${x}^{2}$ from the first group.

${x}^{2} \left(x + 2\right) - \left(x + 2\right)$

Factor out $\left(x + 2\right)$.

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

$\left({x}^{2} - 1\right)$ can be factored as the difference of squares.

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

The complete factorization of $- {x}^{3} - 2 {x}^{2} = - x - 2$ is $\left(x + 2\right) \left(x + 1\right) \left(x - 1\right) = 0$.

Since the factors are equal to zero, we must set each one equal to zero, and solve for $x$.

$x + 2 = 0$
$x = - 2$

$x + 1 = 0$
$x = - 1$

$x - 1 = 0$
$x = 1$

The solution set is $x = \left\{- 2 , - 1 , 1\right\}$