Answer
$(3y+2)(-5y+9)$
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, $
-15y^2+17y+18
,$ the value of $ac$ is $
-15(18)=-270
$ and the value of $b$ is $
17
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-270\}, \{2,-135\}, \{3,-90\}, \{5,-54\}, \{6,-45\}, \{9,-30\}, \{10,-27\}, \{15,-18\},
\{-1,270\}, \{-2,135\}, \{-3,90\}, \{-5,54\}, \{-6,45\}, \{-9,30\}, \{-10,27\}, \{-15,18\},
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-10,27
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
-15y^2-10y+27y+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}
(-15y^2-10y)+(27y+18)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-5y(3y+2)+9(3y+2)
.\end{array}
Factoring the $GCF=
(3y+2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3y+2)(-5y+9)
.\end{array}
