## Intermediate Algebra (12th Edition)

$t=\left\{ 0,4 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $2t^2-8t=0 ,$ 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} 2t(t-4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} 2t=0 \text{ OR } t-4=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} 2t=0 \\\\ t=\dfrac{0}{2} \\\\ t=0 \\\\\text{ OR }\\\\ t-4=0 \\\\ t=4 .\end{array} Hence, the solutions are $t=\left\{ 0,4 \right\} .$ $\bf{\text{Supplementary Solution/s:}}$ In the expression $2t^2-8t ,$ the $GCF$ of the constants of the terms $\{ 2, -8 \}$ 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 $\{ t^2,t \}$ is $t .$ Hence, the entire expression has $GCF= 2t .$ Factoring the $GCF= 2t ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2t \left( \dfrac{2t^2}{2t}-\dfrac{8t}{2t} \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} 2t \left( t^{2-1}-4t^{1-1} \right) \\\\= 2t \left( t^{1}-4t^{0} \right) \\\\= 2t \left( t-4(1) \right) \\\\= 2t \left( t-4 \right) .\end{array}