#### Answer

$ab^2(a-b)(24a+35b)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
24a^3b^2+11a^2b^3-35ab^4
,$ factor first the $GCF.$ Then 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.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms is $GCF=
ab^2
$ since it is the highest expression that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
ab^2(24a^2+11ab-35b^2)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
24(-35)=-840
$ and the value of $b$ is $
11
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-24,35
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
ab^2(24a^2-24ab+35ab-35b^2)
.\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}
ab^2[(24a^2-24ab)+(35ab-35b^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
ab^2[24a(a-b)+35b(a-b)]
.\end{array}
Factoring the $GCF=
(a-b)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
ab^2[(a-b)(24a+35b)]
\\\\=
ab^2(a-b)(24a+35b)
.\end{array}