Question 4357
I am interpreting the problem to be to solve the equation
(((2x^3 = 16x^2}}}

First set the equation equal to zero, by adding {{{ -16x^2}}} to each side.

{{{2x^3 - 16x^2 = 16x^2 -16x^2}}}
{{{2x^3 - 16x^2 = 0}}}


Factor out the common factor which is {{{2x^2}}}

{{{2x^2(x-8) = 0}}}


Set each factor equal to zero:
{{{2x^2 = 0}}} or {{{x-8=0}}}


The first equation {{{2x^2=0}}} has only one solution, x=0.
The second equation {{{x-8=0}}} has one solution, x = 8.


Therefore, the final answer is x= 0 or x= 8.


R^2 at SCC