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

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Answer 1 · 2015-05-24T17:28:31

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

#5c(c^2-9)=0#

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

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

If #c^2-9=0#, then #c^2=9# and #c=+-3#

So, your answers are

#c_1=0#
#c_2=3#
#c_3=-3#