Answer
$\left\{ -2, 2, -3, 3 \right\}
$$
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 equation, $
z^4-13z^2+36=0
,$ then the factored form is
\begin{array}{l}\require{cancel}
z^4-9z^2-4z^2+36=0
\\\\
(z^4-9z^2)-(4z^2-36)=0
\\\\
z^2(z^2-9)-4(z^2-9)=0
\\\\
(z^2-9)(z^2-4)=0
.\end{array}
Equating each factor to zero, then,
\begin{array}{l}\require{cancel}
z^2-9=0
\\\\
z^2=9
\\\\
z=\pm\sqrt{9}
\\\\
z=\pm\sqrt{(3)^2}
\\\\
z=\pm3
,\\\\\text{OR}\\\\
z^2-4=0
\\\\
z^2=4
\\\\
z=\pm\sqrt{4}
\\\\
z=\pm\sqrt{(2)^2}
\\\\
z=\pm2
.\end{array}
Hence, the solutions are $
\left\{ -2, 2, -3, 3 \right\}
$.
