Answer
$-t^2(2t-3)(7t+1)$
Work Step by Step
Factoring the negative $GCF=
-t^2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
-14t^4+19t^3+3t^2
\\\\=
-t^2(14t^2-19t-3)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
-t^2(14t^2-19t-3)
\end{array} has $ac=
14(-3)=-42
$ and $b=
-19
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-21,2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
-t^2(14t^2-21t+2t-3)
.\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^2[(14t^2-21t)+(2t-3)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-t^2[7t(2t-3)+(2t-3)]
.\end{array}
Factoring the $GCF=
(2t-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-t^2[(2t-3)(7t+1)]
\\\\=
-t^2(2t-3)(7t+1)
.\end{array}