## Intermediate Algebra (12th Edition)

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