## Intermediate Algebra (12th Edition)

$2(a+2b)^2 \left( 3-2a-4b+6a^2+24ab+24b^2 \right)$
$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $6(a+2b)^2-4(a+2b)^3+12(a+2b)^4 .$ 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 of the terms $\{ 6,-4,12 \}$ is $2 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ (a+2b)^2,(a+2b)^3,(a+2b)^4 \}$ is $(a+2b)^2 .$ Hence, the entire expression has $GCF= 2(a+2b)^2 .$ Factoring the $GCF= 2(a+2b)^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 2(a+2b)^2 \left( \dfrac{6(a+2b)^2}{2(a+2b)^2}-\dfrac{4(a+2b)^3}{2(a+2b)^2}+\dfrac{12(a+2b)^4}{2(a+2b)^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(a+2b)^2 \left( 3(a+2b)^{2-2}-2(a+2b)^{3-2}+6(a+2b)^{4-2} \right) \\\\= 2(a+2b)^2 \left( 3(a+2b)^{0}-2(a+2b)^{1}+6(a+2b)^{2} \right) \\\\= 2(a+2b)^2 \left( 3(1)-2(a+2b)+6(a+2b)^{2} \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} 2(a+2b)^2 \left( 3(1)-2(a+2b)+6(a^2+4ab+4b^2) \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} 2(a+2b)^2 \left( 3-2a-4b+6a^2+24ab+24b^2 \right) .\end{array}