#### Answer

$(z+2)(z-2)(z^2-5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
z^4-9z^2+20
,$ use the factoring of trinomials in the form $x^2+bx+c.$ Then factor further the resulting expression using the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
In the trinomial expression above, $b=
-9
,\text{ and } c=
20
.$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{
-4,-5
\right\}.$ Using these two numbers, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(z^2-4)(z^2-5)
.\end{array}
The expressions $
z^2
$ and $
4
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
z^2-4
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(z)^2-(2)^2](z^2-5)
\\\\=
[(z+2)(z-2)](z^2-5)
\\\\=
(z+2)(z-2)(z^2-5)
.\end{array}