# How do you factor and solve 5c^3 - 45c = 0?

May 24, 2015

We can see that $5 c$ is multiplying both terms:

$5 c \left({c}^{2} - 9\right) = 0$

Thus, for this equation be true, either $5 c$ and/or $\left({c}^{2} - 9\right)$ have to be equal to zero.

If $5 c = 0$, then $c = 0$.

If ${c}^{2} - 9 = 0$, then ${c}^{2} = 9$ and $c = \pm 3$

${c}_{1} = 0$
${c}_{2} = 3$
${c}_{3} = - 3$