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

2 Answers
mason m
Dec 20, 2015

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)

Topscooter
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 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))

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