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

2 Answers
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)#

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))#