Answer
$10c\left( c-3 \right)\left( c-5 \right)$
Work Step by Step
$10{{c}^{3}}-80{{c}^{2}}+150c$
Factor out 10c:
$10{{c}^{3}}-80{{c}^{2}}+150c$,
$10c\left( {{c}^{2}}-8c+15 \right)$
Find two factors such that their sum is $-8$ and their product is $+15$.
The possibilities are:
$\begin{matrix}
\text{Pairs of factors of 15} & \text{Sums of factors} \\
3,5 & 8 \\
-3,-5 & -8 \\
1,15 & 16 \\
-1,-15 & -16 \\
\end{matrix}$
Here, we see that -3 and -5 are the correct options based on the description above.
$\begin{align}
& \left( -3 \right)+\left( -5 \right)=-8 \\
& \left( -3 \right)\cdot \left( -5 \right)=15
\end{align}$
Check:
Multiply the factors and simplify:
$\begin{align}
& 10c\left( c-3 \right)\left( c-5 \right)=10c\left( c\left( c-5 \right)-3\left( c-5 \right) \right) \\
& =10c\left( {{c}^{2}}-5c-3c+15 \right) \\
& =10c\left( {{c}^{2}}-8c+15 \right) \\
& =10{{c}^{3}}-80c+150
\end{align}$
Thus, the answer checks.
