Answer
$(3p+4)(2p+5)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6p^2+23p+20
,$ 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.
$\bf{\text{Solution Details:}}$
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
6(20)=120
$ and the value of $b$ is $
23
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
8,15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6p^2+8p+15p+20
.\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}
(6p^2+8p)+(15p+20)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2p(3p+4)+5(3p+4)
.\end{array}
Factoring the $GCF=
(3p+4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3p+4)(2p+5)
.\end{array}
