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

May 25, 2017

${x}^{4} - 13 {x}^{2} + 4 = \left(x - \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x - \frac{3}{2} + \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{17}}{2}\right)$

#### Explanation:

This quartic can be factored as a "quadratic in ${x}^{2}$" then taking square roots, but the resulting form of the answer is not immediately simple.

Alternatively, consider the following:

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

So putting $k = \sqrt{17}$, we find:

$\left({x}^{2} - \sqrt{17} x + 2\right) \left({x}^{2} + \sqrt{17} x + 2\right) = {x}^{4} - 13 {x}^{2} + 4$

Using the quadratic formula, the zeros of ${x}^{2} \pm \sqrt{17} x + 2$ are:

$\frac{\pm \sqrt{17} \pm \sqrt{17 - 4 \left(1\right) \left(2\right)}}{2 \cdot 1} = \pm \frac{\sqrt{17}}{2} \pm \frac{\sqrt{9}}{2} = \pm \frac{\sqrt{17}}{2} \pm \frac{3}{2}$

So we find:

${x}^{4} - 13 {x}^{2} + 4 = \left(x - \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x - \frac{3}{2} + \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} - \frac{\sqrt{17}}{2}\right) \left(x + \frac{3}{2} + \frac{\sqrt{17}}{2}\right)$

Alternatively again, consider the following:

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

So putting $h = 3$, we find:

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

Using the quadratic formula, the zeros of ${x}^{2} \pm 3 x - 2$ are:

$\frac{\pm 3 \pm \sqrt{9 - 4 \left(1\right) \left(- 2\right)}}{2 \cdot 1} = \pm \frac{3}{2} \pm \frac{\sqrt{17}}{2}$

Hence the same linear factorisation as before.