Answer
$t(t-3)(5t-6)$
Work Step by Step
Factoring the $GCF=
t
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
5t^3-21t^2+18t
\\\\=
t(5t^2-21t+18)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
t(5t^2-21t+18)
\end{array} has $ac=
5(18)=90
$ and $b=
-21
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-15,-6
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
t(5t^2-15t-6t+18)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
t[(5t^2-15t)-(6t-18)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
t[5t(t-3)-6(t-3)]
.\end{array}
Factoring the $GCF=
(t-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
t[(t-3)(5t-6)]
\\\\=
t(t-3)(5t-6)
.\end{array}
