# How do you find the roots of x^3-12x+16=0?

May 29, 2017

$x = 2 \text{ with multiplicity 2}$
$x = - 4 \text{ with multiplicity 1}$

#### Explanation:

$\text{note that } {2}^{3} - 12 \left(2\right) + 16 = 0$

$\Rightarrow x = 2 \text{ is a root and " (x-2)" is a factor}$

$\Rightarrow {x}^{2} \left(x - 2\right) + 2 x \left(x - 2\right) - 8 \left(x - 2\right)$

$= \left(x - 2\right) \left({x}^{2} + 2 x - 8\right)$

$= \left(x - 2\right) \left(x + 4\right) \left(x - 2\right)$

$\Rightarrow {\left(x - 2\right)}^{2} \left(x + 4\right) = 0$

$\Rightarrow x = 2 \text{ multiplicity of 2}$

$\text{and " x=-4" multiplicity of 1}$

May 29, 2017

The roots are $4$ and $2$(double root)

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 12 x + 16$

Then,

We see that

$f \left(2\right) = {2}^{3} - 12 \cdot 2 + 16 = 8 - 24 + 16 = 0$

Therefore,

$\left(x - 2\right)$ is a factor of $f \left(x\right)$

To find the other factors, we perform a long division

$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a}$$|$${x}^{3} + 0 {x}^{2} - 12 x + 16$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} + 2 x - 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$${x}^{3} - 2 {x}^{2}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$0 + 2 {x}^{2} - 12 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$+ 2 {x}^{2} - 4 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$+ 0 - 8 x + 16$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a}$$- 8 x + 16$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a}$$- 0 + 0$

The quotient is

${x}^{2} + 2 x - 8 = \left(x - 2\right) \left(x + 4\right)$

Therefore,

${x}^{3} - 12 x + 16 = {\left(x - 2\right)}^{2} \left(x + 4\right)$ graph{x^3-12x+16 [-16.15, 15.88, -3.97, 12.05]}