## Intermediate Algebra (12th Edition)

$x=\left\{ 0,9 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $3x^2-27x=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} 3x(x-9)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} 3x=0 \text{ OR } x-9=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} 3x=0 \\\\ x=\dfrac{0}{3} \\\\ x=0 \\\\\text{ OR }\\\\ x-9=0 \\\\ x=9 .\end{array} Hence, the solutions are $x=\left\{ 0,9 \right\} .$ $\bf{\text{Supplementary Solution/s:}}$ In the expression $3x^2-27x ,$ the $GCF$ of the constants of the terms $\{ 3,-27 \}$ is $3 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ x^2,x \}$ is $x .$ Hence, the entire expression has $GCF= 3x .$ Factoring the $GCF= 3x ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3x \left( \dfrac{3x^2}{3x}-\dfrac{27x}{3x} \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} 3x \left( x^{2-1}-9x^{1-1} \right) \\\\= 3x \left( x^{1}-9x^{0} \right) \\\\= 3x \left( x-9(1) \right) \\\\= 3x \left( x-9 \right) .\end{array}