# How do you solve 0=4x^3-20x+16?

Apr 30, 2016

$\left\{\begin{matrix}x = 1 \\ x = - \frac{1}{2} + \frac{\sqrt{17}}{2} \\ x = - \frac{1}{2} - \frac{\sqrt{17}}{2}\end{matrix}\right.$

#### Explanation:

First note that all of the terms are divisible by $4$, so we can separate that out to get:

$0 = 4 {x}^{3} - 20 x + 16 = 4 \left({x}^{3} - 5 x + 4\right)$

Next note that the sum of the coefficients of the simplified cubic is zero. that is:$1 - 5 + 4 = 0$. Hence $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} - 5 x + 4 = \left(x - 1\right) \left({x}^{2} + x - 4\right)$

The remaining quadratic factor is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 1$ and $c = - 4$. This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \cdot 1 \cdot - 4\right)}}{2 \cdot 1}$

$= \frac{- 1 \pm \sqrt{17}}{2} = - \frac{1}{2} \pm \frac{\sqrt{17}}{2}$