Intermediate Algebra (12th Edition)

$(2u+v)$
$\bf{\text{Solution Outline:}}$ 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 $(2u+v) .$