Question 87059
Solve for m:
{{{m^3+2m^2-3m = 0}}} First, notice that m is common to all three terms, so you can factor out an m.
{{{m(m^2+2m-3) = 0}}} Apply the zero products principle: If a*b = 0, then either a = 0, or b = 0, or both a and b = 0.  In this case:
{{{m = 0}}} or {{{m^2+2m-3 = 0}}} So m = 0 is one of the solutions.
Factor the quadratic:
{{{m^2+2m-3 = (m+3)(m-1)}}} So...
{{{(m+3)(m-1) = 0}}} Apply the zero products principle again:
{{{m+3 = 0}}} or {{{m-1 = 0}}}
The other two solutions are:
m = -3
m = 1
The three solutions are:
m = 0
m = 1
m = -3