Answer
$(4x-3)(6x-5)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
24x^2-38x+15
,$ 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 $
24(15)=360
$ and the value of $b$ is $
-38
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-18,-20
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
24x^2-18x-20x+15
.\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}
(24x^2-18x)-(20x-15)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
6x(4x-3)-5(4x-3)
.\end{array}
Factoring the $GCF=
(4x-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(4x-3)(6x-5)
.\end{array}
