Answer
$7ab\left( 2a^{2}b+a-3a^{4}b^{2}+6b^{3} \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
14a^3b^2+7a^2b-21a^5b^3+42ab^4
.$ 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 $(
14,7,-21,42
)$ is $
7
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $(
a^3b^2,a^2b,a^5b^3,ab^4
)$ is $
ab
.$ Hence, the entire expression has $
GCF=7ab
.$
Factoring the $
GCF=7ab
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
7ab\left( \dfrac{14a^3b^2}{7ab}+\dfrac{7a^2b}{7ab}-\dfrac{21a^5b^3}{7ab}+\dfrac{42ab^4}{7ab} \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}
7ab\left( 2a^{3-1}b^{2-1}+a^{2-1}b^{1-1}-3a^{5-1}b^{3-1}+6a^{1-1}b^{4-1} \right)
\\\\=
7ab\left( 2a^{2}b^{1}+a^{1}b^{0}-3a^{4}b^{2}+6a^{0}b^{3} \right)
\\\\=
7ab\left( 2a^{2}b+a(1)-3a^{4}b^{2}+6(1)b^{3} \right)
\\\\=
7ab\left( 2a^{2}b+a-3a^{4}b^{2}+6b^{3} \right)
.\end{array}
