#### Answer

$x=2,-5,7$

#### Work Step by Step

Given: $x(x-2)^3-35(x-2)^2=0$
Need to use factorization to solve the quadratic equation.
This can be factorized as follows:
$(x-2)^2(x^2-2x-35)=0$
Now $(x-2)^2=0$ and $(x^2-2x-35)=0$
This implies $x=2$
and $(x^2-2x-35)=0$
or, $x^2+5x-7x-35=0$
or, $(x-7)(x+5)=0$
This implies $x=7,-5$
Hence, $x=2,-5,7$