Answer
$(5x-6)(2x+3)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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.
$\bf{\text{Solution Details:}}$
In the given expression, $
10x^2+3x-18
,$ the value of $ac$ is $
10(-18)=-180
$ and the value of $b$ is $
3
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-180\}, \{2,-90\}, \{3,-60\}, \{4,-45\}, \{5,-36\}, \{6,-30\}, \{9,-20\}, \{10,-18\}, \{12,-15\},
\{-1,180\}, \{-2,90\}, \{-3,60\}, \{-4,45\}, \{-5,36\}, \{-6,30\}, \{-9,20\}, \{-10,18\}, \{-12,15\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-12,15
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
10x^2-12x+15x-18
.\end{array}
Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(10x^2-12x)+(15x-18)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2x(5x-6)+3(5x-6)
.\end{array}
Factoring the $GCF=
(5x-6)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(5x-6)(2x+3)
.\end{array}