# How do you solve x^3 + 4x^2 - 20x = 80?

##### 1 Answer
Nov 28, 2015

Subtract $80$ from both sides, then factor by grouping to find:

$x = \pm 2 \sqrt{5}$ or $x = - 4$

#### Explanation:

First subtract $80$ from both sides to get:

${x}^{3} + 4 {x}^{2} - 20 x - 80 = 0$

Factor the left hand side by grouping:

${x}^{3} + 4 {x}^{2} - 20 x - 80$

$= \left({x}^{3} + 4 {x}^{2}\right) - \left(20 x + 80\right)$

$= {x}^{2} \left(x + 4\right) - 20 \left(x + 4\right)$

$= \left({x}^{2} - 20\right) \left(x + 4\right)$

$= \left(x - \sqrt{20}\right) \left(x + \sqrt{20}\right) \left(x + 4\right)$

$= \left(x - \sqrt{{2}^{2} \cdot 5}\right) \left(x + \sqrt{{2}^{2} \cdot 5}\right) \left(x + 4\right)$

$= \left(x - 2 \sqrt{5}\right) \left(x + 2 \sqrt{5}\right) \left(x + 4\right)$

So the roots of ${x}^{3} + 4 {x}^{2} - 20 x = 80$ are $x = \pm 2 \sqrt{5}$ and $x = - 4$