Answer
$3(2x+1)(4x+5)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, factor the GCF of the terms. Then, to factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Using the $GCF=
3
,$ the given expression, $
24x^2+42x+15
,$ is equivalent to
\begin{array}{l}\require{cancel}
3(8x^2+14x+5)
.\end{array}
In the expression above, the value of $ac$ is $
8(5)=40
$ and the value of $b$ is $
14
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,40\}, \{2,20\}, \{4,10\}, \{5,8\},
\{-1,-40\}, \{-2,-20\}, \{-4,-10\}, \{-5,-8\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
4,10
\}.$ Using these $2$ numbers to decompose the middle term of the expression above results to
\begin{array}{l}\require{cancel}
3(8x^2+4x+10x+5)
.\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}
3[(8x^2+4x)+(10x+5)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3[4x(2x+1)+5(2x+1)]
.\end{array}
Factoring the $GCF=
(2x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
3[(2x+1)(4x+5)]
\\\\=
3(2x+1)(4x+5)
.\end{array}
