Question 403036
{{{7y^3-28y=0}}}
We start by factoring. And factoring starts with the Greastest Common Facotr (GCF). The GCF here is 7y:
{{{7y(y^2-4)=0}}}
The second factor is a difference of squares so we can use the pattern {{{a^2-b^2=(a+b)(a-b)}}} with "a" being y and "b" being 2:
7y(y+2)(y-2) = 0
From the Zero Product Property we know that this (or any) product can be zero <i>only</i> if one (or more) of the factors is zero. So:
7y = 0 or y+2 = 0 or y-2 = 0
Solving each of these we get:
y = 0 or y = -2 or y = 2
These are the three solutions to your equation.