#### Answer

$3b(b+1)(2b-5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6b^3-9b^2-15b
,$ factor first the $GCF.$ Then factor the resulting trinomial by finding 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.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
6,-9,-15
\}$ is $
3
$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
b^3,b^2,b
\}$ is $
b
.$ Hence, the entire expression has $GCF=
3b
.$
Factoring the $GCF=
3b
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
3b \left( \dfrac{6b^3}{3b}-\dfrac{9b^2}{3b}-\dfrac{15b}{3b} \right)
\\\\=
3b(2b^2-3b-5)
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
2(-5)=-10
$ and the value of $b$ is $
-3
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-10\}, \{2,-5\},
\\
\{-1,10\}, \{-2,5\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
2,-5
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3b(2b^2+2b-5b-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}
3b[(2b^2+2b)-(5b+5)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3b[2b(b+1)-5(b+1)]
.\end{array}
Factoring the $GCF=
(b+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
3b[(b+1)(2b-5)]
\\\\=
3b(b+1)(2b-5)
.\end{array}