Answer
$(2u+v)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
12u^2+10uv+2v^2
,$ factor first the $GCF.$ Then, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
2(6u^2+5uv+v^2)
.\end{array}
In the trinomial expression above the value of $ac$ is $
6(1)=6
$ and the value of $b$ is $
5
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,6\}, \{2,3\},
\{-1,-6\}, \{-2,-3\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
2,3
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2(6u^2+2uv+3uv+v^2)
.\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}
2[(6u^2+2uv)+(3uv+v^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2[2u(3u+v)+v(3u+v)]
.\end{array}
Factoring the $GCF=
(3u+v)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2[(3u+v)(2u+v)]
\\\\=
2(3u+v)(2u+v)
.\end{array}
The missing factor of the given expression is $
(2u+v)
.$