Question 510433
use factoring to solve the quadratic equation 
x(x-2)^3 -35(x-2)^2=0
(x-2)^2[x(x-2)-35]=0
(x-2)^2=0
x=2 (multiplicity of 2)
..
[x(x-2)-35]=0
x^2-2x-35=0
(x-7)(x+5)=0
x-7=0
x=7
or
x+5=0
x=-5
ans:
zeros or roots are: -5, 2, 2, and 7