Question 131936
I presume you meant {{{w^4 - 8w^2 - 9}}}


Let {{{x=w^2}}}, then substitute:


{{{x^2-8x-9}}}


{{{-9*1=-9}}} and {{{-9+1=-8}}}, so the factors of {{{x^2-8x-9}}} are:


{{{(x-9)(x+1)}}}


but {{{x=w^2}}}, so two of the factors of {{{w^4 - 8w^2 - 9}}} are:


{{{(w^2-9)(w^2+1)}}}


The first of these factors is the difference of two squares and factors directly to:


{{{(w^2-9)=(w-3)(w+3)}}}


For the other factor, we need to use the fact that {{{a}}} is a zero of a single variable polynomial in x if and only if {{{x-a}}} is a factor of the polynomial.


So, solve {{{w^2+1=0}}}


{{{w^2=-1}}} => {{{w = i}}} or {{{w = -i}}} where {{{i^2=-1}}}, therefore the two factors of {{{w^2+1}}} are {{{(w-i)(w+i)}}}


In summary:
{{{w^4 - 8w^2 - 9=(w-3)(w+3)(w-i)(w+i)}}}


If you were only supposed to factor over the real numbers, then your answer would be:
{{{w^4 - 8w^2 - 9=(w-3)(w+3)(w^2+1)}}}