Answer
$20z(2z+1)\left( 3z+4 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
15(2z+1)^3+10(2z+1)^2-25(2z+1)
.$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
15,10,-25
\}$ is $
5
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{
(2z+1)^3,(2z+1)^2,(2z+1)
\}$ is $
(2z+1)
.$ Hence, the entire expression has $GCF=
5(2z+1)
.$
Factoring the $GCF=
5(2z+1)
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
5(2z+1) \left( \dfrac{15(2z+1)^3}{5(2z+1)}+\dfrac{10(2z+1)^2}{5(2z+1)}-\dfrac{25(2z+1)}{5(2z+1)} \right)
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
5(2z+1) \left( 3(2z+1)^{3-1}+2(2z+1)^{2-1}-5(2z+1)^{1-1} \right)
\\\\=
5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)^{1}-5(2z+1)^{0} \right)
\\\\=
5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)-5(1) \right)
\\\\=
5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)-5 \right)
.\end{array}
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the expression above is
\begin{array}{l}\require{cancel}
5(2z+1) \left( 3(4z^2+4z+1)+2(2z+1)-5 \right)
.\end{array}
Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
5(2z+1) \left( 12z^2+12z+3+4z+2-5 \right)
\\\\=
5(2z+1) \left( 12z^2+(12z+4z)+(3+2-5) \right)
\\\\=
5(2z+1) \left( 12z^2+16z \right)
.\end{array}
Factoring the $GCF=4z$ from the last factor of the expression above results to
\begin{array}{l}\require{cancel}
5(2z+1) 4z\left( 3z+4 \right)
\\\\=
20z(2z+1)\left( 3z+4 \right)
.\end{array}