#### Answer

$-(x+3)(x-3)(x^2+2)$

#### Work Step by Step

Factoring the negative $GCF=
-1
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
-x^4+7x^2+18
\\\\=
-(x^4-7x^2-18)
.\end{array}
Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
(x^4-7x^2-18)
\end{array} has $c=
-18
$ and $b=
-7
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-9,2
\right\}.$ Using these two numbers, the $\text{
expression
}$ above is equivalent to
\begin{array}{l}\require{cancel}
-(x^2-9)(x^2+2)
.\end{array}
The expressions $
x^2
$ and $
9
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-9
,$ 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}
-[(x)^2-(3)^2](x^2+2)
\\\\=
-(x+3)(x-3)(x^2+2)
.\end{array}