# How do you factor 2x^4-2x^2-40?

Dec 20, 2015

$2 \left({x}^{2} - 5\right) \left({x}^{2} + 4\right)$

#### Explanation:

Factor out a $2$.

$= 2 \left({x}^{4} - {x}^{2} - 20\right)$

Now, to make this look more familiar, say that $u = {x}^{2}$.

$= 2 \left({u}^{2} - u - 20\right)$

Which can be factorized as follows:

$= 2 \left(u - 5\right) \left(u + 4\right)$

Plug ${x}^{2}$ back in for $u$.

$= 2 \left({x}^{2} - 5\right) \left({x}^{2} + 4\right)$

${x}^{2} - 5$ can optionally be treated as a difference of squares.

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

Dec 20, 2015

You change the variable, and the result is 2(x - sqrt(2+isqrt(316))/2)(x + sqrt(2+isqrt(316))/2))(x - sqrt(2-isqrt(316))/2))(x + sqrt(2-isqrt(316))/2))

#### Explanation:

This is quite a remarkable polynomial here, it only has even powers! So we can change the variable, let's say $X = {x}^{2}$.

So we now have to factorise $2 {X}^{2} - 2 X + 40$, which is pretty easy with the quadratic formula.

$\Delta = {b}^{2} - 4 a c = 4 - 4 \cdot 2 \cdot 40 = - 316$. This polynomial has complex roots only.

${X}_{1} = \frac{2 - i \sqrt{316}}{4} =$ and ${X}_{2} = \frac{2 + i \sqrt{316}}{4}$.

$2 {X}^{2} - 2 X + 40 = 2 \left(X - \frac{2 + i \sqrt{316}}{4}\right) \left(X - \frac{2 - i \sqrt{316}}{4}\right)$. But $X = {x}^{2}$ so $2 {x}^{4} - 2 {x}^{2} + 40 = 2 \left({x}^{2} - \frac{2 + i \sqrt{316}}{4}\right) \left({x}^{2} - \frac{2 - i \sqrt{316}}{4}\right)$

So finally, you can factorize it as 2(x - sqrt(2+isqrt(316))/2)(x + sqrt(2+isqrt(316))/2))(x - sqrt(2-isqrt(316))/2))(x + sqrt(2-isqrt(316))/2))