Question 536920
I would expect that this problem was meant to say factor {{{x^4-13x^2+40}}}
In that case, since you know that {{{y^2-13y+40=(y-8)(y-5)}}}
your first step would be
{{{x^4-13x^2+40=(x^2-8)(x^2-5)}}}
From there on it does not get too pretty
{{{(x^2-5)=(x+sqrt(5))(x-sqrt(5))}}} and
{{{(x^2-8)=(x+sqrt(8))(x-sqrt(8))=(x+2sqrt(2))(x-2sqrt(2))}}}
So you could say
{{{x^4-13x^2+40=(x^2-8)(x^2-5)=(x+2sqrt(2))(x-2sqrt(2))(x+sqrt(5))(x-sqrt(5))}}}
However it gets much uglier with {{{x^4-13x^2+4}}}.
The solutions to {{{y^2-13y+4=0}}}
are {{{y = (13 +- sqrt( 13^2-4*1*4 ))/(2*1)=(13 +- sqrt(153))/2 }}}
They are both positive, so we would end up with
{{{x^4-13x^2+4=(x^2-(13 - sqrt(153))/2)(x^2-(13 + sqrt(153))/2)=(x+sqrt((13 - sqrt(153))/2))(x-sqrt((13 - sqrt(153))/2))(x+sqrt((13 + sqrt(153))/2))(x-sqrt((13 + sqrt(153))/2))}}}