# How do you find the roots of x^3-18x+27=0?

Oct 31, 2016

The roots are $\left(3 , - \frac{3}{2} + 3 \frac{\sqrt{5}}{2} , - \frac{3}{2} - 3 \frac{\sqrt{5}}{2}\right)$

#### Explanation:

Let $f \left(x\right) = {x}^{3} - 18 x + 27$
Then by trial and error, $f \left(3\right) = 27 - 54 + 27 = 0$
so $x = 3$ is a root
Then we do a long division
${x}^{3}$$\textcolor{w h i t e}{a a a a a}$$- 18 x + 27$∣$x - 3$
${x}^{3} - 3 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a}$∣${x}^{2} + 3 x - 9$
$0 + 3 {x}^{2} - 18 x$
$\textcolor{w h i t e}{a a a a}$$0 - 9 x$
$\textcolor{w h i t e}{a a a a a a}$$- 9 x + 27$
$\textcolor{w h i t e}{a a a a a a}$$- 9 x + 27$
$\textcolor{w h i t e}{a a a a a a a a}$$0 + 0$
To find the other roots, we must solve
${x}^{x} + 3 x - 9 = {x}^{2} + 3 x + \frac{9}{4} - 9 - \frac{9}{4}$
${\left(x + \frac{3}{2}\right)}^{2} - \frac{45}{4}$
So ${\left(x + \frac{3}{2}\right)}^{2} = \frac{45}{4}$$\implies$$x + \frac{3}{2} = \pm \frac{\sqrt{45}}{2}$
$x = - \frac{3}{2} \pm 3 \frac{\sqrt{5}}{2}$