Answer
$x=\left\{ -\dfrac{7}{2},0,4 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
2x^3-x^2-28x=0
,$ express it in the factored form. Then equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=x$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(2x^2-x-28)=0
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
2(-28)=-56
$ and the value of $b$ is $
-1
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,-56 \}, \{ 2,-28 \}, \{ 4,-14 \}, \{ 7,-8\},
\\
\{ -1,56 \}, \{ -2,28 \}, \{ -4,14 \}, \{ -7,8\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
7,-8
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x(2x^2+7x-8x-28)=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
x[(2x^2+7x)-(8x+28)]=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x[x(2x+7)-4(2x+7)]=0
.\end{array}
Factoring the $GCF=
(2x+7)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x[(2x+7)(x-4)]=0
\\\\
x(2x+7)(x-4)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions to the equation above are
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
2x+7=0
\\\\\text{OR}\\\\
x-4=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
2x+7=0
\\\\
2x=-7
\\\\
x=-\dfrac{7}{2}
\\\\\text{OR}\\\\
x-4=0
\\\\
x=4
.\end{array}
Hence, $
x=\left\{ -\dfrac{7}{2},0,4 \right\}
.$
