Answer
$7(y-1)(2y+3)$
Work Step by Step
Factoring the $GCF=
7
,$ the given expression, $
14y^2+7y-21
,$ is equivalent to
\begin{align*}
7(2y^2+y-3)
\end{align*}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression above has $ac=
2(-3)=-6
$ and $b=
1
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-2,3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{align*}
7(2y^2-2y+3y-3)
\end{align*}
Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to
\begin{align*}
7[(2y^2-2y)+(3y-3)]
\end{align*}
Factoring the $GCF$ in each group results to
\begin{align*}
7[2y(y-1)+3(y-1)]
\end{align*}
Factoring the $GCF=
(y-1)
$ of the entire expression above results to
\begin{align*}
&
7[(y-1)(2y+3)]
\\&=
7(y-1)(2y+3)
\end{align*}