Answer
$3x(x^{2}+2)(x-3)(x+3)$
Work Step by Step
$ 3x^{5}-21x^{3}-54x\qquad$...factor out the common term, $3x$.
$=3x(x^{4}-7x^{2}-18)$
... Searching for two factors of $ac=-18$ whose sum is $b=-7,$
we find$\qquad 2$ and $-9.$
Rewrite the middle term and factor in pairs:
$=3x(x^{4}+2x^{2}-9x^{2}-18)=$
$=3x[x^{2}(x^{2}+2)-9(x^{2}+2)]$
$=3x(x^{2}+2)(x^{2}-9)\qquad$...recognize the difference of two squares: $a^{2}-b^{2}=(a-b)(a+b)$
=$3x(x^{2}+2)(x-3)(x+3)$
