How do you factor and solve $5 c^{3} - 45 c = 0$ $5c^3 - 45c = 0$ ?

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Answer 1 · 2015-05-24T17:28:31

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

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

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

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

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

So, your answers are

$c_{1} = 0$ $c_1=0$
$c_{2} = 3$ $c_2=3$
$c_{3} = - 3$ $c_3=-3$

How do you factor and solve 5c^3 - 45c = 05c3−45c=05c^3 - 45c = 0?