Question

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

Answer 1

`2(x^2-5)(x^2+4)`

Explanation:

Factor out a `2`.

`=2(x^4-x^2-20)`

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

`=2(u^2-u-20)`

Which can be factorized as follows:

`=2(u-5)(u+4)`

Plug `x^2` back in for `u`.

`=2(x^2-5)(x^2+4)`

`x^2-5` can optionally be treated as a difference of squares.

`=2(x+sqrt5)(x-sqrt5)(x^2+4)`

Answer 2

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 `2X^2 - 2X + 40`, which is pretty easy with the quadratic formula.

`Delta = b^2 - 4ac = 4 - 4*2*40 = -316`. This polynomial has complex roots only.

`X_1 = (2 - isqrt(316))/4 =` and `X_2 = (2+isqrt(316))/4`.

`2X^2 - 2X + 40 = 2(X - (2+isqrt316)/4)(X - (2-isqrt316)/4)`. But `X=x^2` so `2x^4 - 2x^2 + 40 = 2(x^2 - (2+isqrt316)/4)(x^2 - (2-isqrt316)/4)`

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))`