Answer
$t=\left\{ 0,4 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
