# How do you factor completely h^3-27h+10?

May 9, 2017

${h}^{3} - 27 h + 10 = \left(h - 5\right) \left({h}^{2} + 5 h - 2\right) = \left(h - 5\right) \left(h + \frac{5}{2} + \frac{\sqrt{33}}{2}\right) \left(h + \frac{5}{2} - \frac{\sqrt{33}}{2}\right)$

#### Explanation:

As the function is ${h}^{3} - 27 h + 10$, one of the zeros could be factor of $10$ i.e. $\pm 2$ or $\pm 5$. As is seen $5$ is a zero of the function as

${\left(5\right)}^{3} - 27 \left(5\right) + 10 = 125 - 135 + 10 = 0$

and hence $\left(h - 5\right)$ is a factor of ${h}^{3} - 27 h + 10$

Dividing it by $\left(h - 5\right)$, we get

${h}^{2} \left(h - 5\right) + 5 h \left(h - 5\right) - 2 \left(h - 5\right) = \left(h - 5\right) \left({h}^{2} + 5 h - 2\right)$

As discriminant of ${h}^{2} + 5 h - 2$ is ${5}^{2} - 4 \times 1 \times \left(- 2\right) = 33$, which is not a perfect square and hence we can only have additional irrational factor.

${h}^{2} + 5 h - 2 = \left({h}^{2} + 2 \times \frac{5}{2} \times h + {\left(\frac{5}{2}\right)}^{2}\right) - {\left(\frac{5}{2}\right)}^{2} - 2$

= ${\left(h + \frac{5}{2}\right)}^{2} - \frac{33}{4} = {\left(h + \frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{33}}{2}\right)}^{2}$

= $\left(h + \frac{5}{2} + \frac{\sqrt{33}}{2}\right) \left(h + \frac{5}{2} - \frac{\sqrt{33}}{2}\right)$

Hence ${h}^{3} - 27 h + 10 = \left(h - 5\right) \left(h + \frac{5}{2} + \frac{\sqrt{33}}{2}\right) \left(h + \frac{5}{2} - \frac{\sqrt{33}}{2}\right)$