How do you solve #12x^2=48x#?

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Answer 1 · 2018-06-25T19:53:04

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Rearrange:

#12x^2-48x =0 #

Factor out the #x#:

#x(12x-48) = 0 #

Hence product of two polynomials equaling 0 means

#x = 0 #

and

#12x - 48 = 0 #

#12x = 48 => x= 48/12 #

#x = 0 # and # x= 4 #