Answer
The solution set is $\left\{0, \dfrac{4}{3}, \dfrac{7}{5}\right\}$.
Work Step by Step
Factr out $2x$ to obtain:
$$2x(15x^2-41x+28x)=0$$
Factor the trinomial to obtain:
$$2x(5x-7)(3x-4)=0$$
Use the Zero-Product Property by equating each factor to zero
Then, solve each equation to obtain:
\begin{align*}
2x&=0 &\text{or}& &5x-7=0& &\text{or}& &3x-4=0\\.
x&=0 &\text{or}& &5x=7& &\text{or}& &3x=4\\
x&=0 &\text{or}& &x=\frac{7}{5}& &\text{or}& &x=\frac{4}{3}\\
\end{align*}
Checking:
\begin{align*}
30(0^3)-82(0^2)+56(0)&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
\begin{align*}
30\left(\frac{7}{5}\right)^3-82\left(\frac{7}{5}\right)+56\left(\frac{7}{5}\right)&=0\\
30\left(\frac{343}{125}\right)-82\left(\frac{49}{25}\right)+\frac{392}{5}&=0\\
82.32-160.72+78.4&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
\begin{align*}
30\left(\frac{4}{3}\right)^3-82\left(\frac{4}{3}\right)+56\left(\frac{4}{3}\right)&=0\\
30\left(\frac{64}{27}\right)-82\left(\frac{16}{9}\right)+\frac{224}{3}&=0\\
\frac{640}{9}-\frac{1312}{9}+\frac{672}{9}&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
Therefore, the solution set is $\left\{0, \dfrac{4}{3}, \dfrac{7}{5}\right\}$.
