# How do you factor and solve each polynomial completely y=x^4-14x^2+40?

Jan 8, 2017

${x}^{4} - 14 {x}^{2} + 40 = \left(x - \sqrt{10}\right) \left(x + \sqrt{10}\right) \left(x - 2\right) \left(x + 2\right)$

with zeros: $\pm \sqrt{10}$ and $\pm 2$

#### Explanation:

The difference of squares identity can be written:

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

We use this a couple of times below.

Note that $10 + 4 = 14$ and $10 \cdot 4 = 40$

So we find:

${x}^{4} - 14 {x}^{2} + 40 = \left({x}^{2} - 10\right) \left({x}^{2} - 4\right)$

$\textcolor{w h i t e}{{x}^{4} - 14 {x}^{2} + 40} = \left({x}^{2} - {\left(\sqrt{10}\right)}^{2}\right) \left({x}^{2} - {2}^{2}\right)$

$\textcolor{w h i t e}{{x}^{4} - 14 {x}^{2} + 40} = \left(x - \sqrt{10}\right) \left(x + \sqrt{10}\right) \left(x - 2\right) \left(x + 2\right)$

So this quartic has zeros:

$x = \pm \sqrt{10} \text{ }$ and $\text{ } x = \pm 2$