Answer
$(z+3)(z-3)(z+2)(z-2)$
Work Step by Step
The 2 numbers whose product is $ac=
1(36)=36
$ and whose sum is $b=
-13
$ are $\{
-9,-4
.\}$ Using these two numbers to decompose the middle term of the given expression, $
z^4-13z^2+36
,$ then the factored form is
\begin{array}{l}\require{cancel}
z^4-9z^2-4z^2+36
\\\\=
(z^4-9z^2)-(4z^2-36)
\\\\=
z^2(z^2-9)-4(z^2-9)
\\\\=
(z^2-9)(z^2-4)
.\end{array}
Using $a^2-b^2=(a+b)(a-b)$, then the complete factored form of the equation above is
\begin{array}{l}\require{cancel}
(z+3)(z-3)(z+2)(z-2)
.\end{array}
