# How do you factor z^3-z^2-13z+4?

Apr 14, 2016

${z}^{3} - {z}^{2} - 13 z + 4 = \left(z - 4\right) \left(z + \frac{3}{2} - \frac{\sqrt{5}}{2}\right) \left(z + \frac{3}{2} + \frac{\sqrt{5}}{2}\right)$

#### Explanation:

By the rational root theorem, any rational zeros must be expressible in the form $\frac{p}{q}$ for some integers $p$ and $q$ with $p$ a divisor of the constant term $4$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$

Let $f \left(z\right) = {z}^{3} - {z}^{2} - 13 z + 4$

We find $f \left(4\right) = 64 - 16 - 52 + 4 = 0$, so $z = 4$ is a zero and $\left(z - 4\right)$ is a factor:

${z}^{3} - {z}^{2} - 13 z + 4 = \left(z - 4\right) \left({z}^{2} + 3 z - 1\right)$

We can factor the remaining quadratic factor by completing the square:

${z}^{2} + 3 z - 1$

$= {\left(z + \frac{3}{2}\right)}^{2} - \frac{9}{4} + 1$

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

$= {\left(z + \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2}$

$= \left(z + \frac{3}{2} - \frac{\sqrt{5}}{2}\right) \left(z + \frac{3}{2} + \frac{\sqrt{5}}{2}\right)$

Putting it all together:

${z}^{3} - {z}^{2} - 13 z + 4 = \left(z - 4\right) \left(z + \frac{3}{2} - \frac{\sqrt{5}}{2}\right) \left(z + \frac{3}{2} + \frac{\sqrt{5}}{2}\right)$