# How do you factor z^4 - 10z^2 + 9?

Sep 2, 2016

${z}^{4} - 10 {z}^{2} + 9 = \left(z - 1\right) \left(z + 1\right) \left(z - 3\right) \left(z + 3\right)$

#### Explanation:

The difference of square identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this a couple of times.

Let us treat this as a quadratic in ${z}^{2}$. Note that the sum of the coefficients is $0$. That is $1 - 10 + 9 = 0$. Hence $\left(\left({z}^{2}\right) - 1\right)$ is a factor and the other factor must be $\left(\left({z}^{2}\right) - 9\right)$...

${z}^{4} - 10 {z}^{2} + 9 = {\left({z}^{2}\right)}^{2} - 10 \left({z}^{2}\right) - 9$

$\textcolor{w h i t e}{{z}^{4} - 10 {z}^{2} + 9} = \left({z}^{2} - 1\right) \left({z}^{2} - 9\right)$

$\textcolor{w h i t e}{{z}^{4} - 10 {z}^{2} + 9} = \left({z}^{2} - {1}^{2}\right) \left({z}^{2} - {3}^{2}\right)$

$\textcolor{w h i t e}{{z}^{4} - 10 {z}^{2} + 9} = \left(z - 1\right) \left(z + 1\right) \left(z - 3\right) \left(z + 3\right)$