Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 52

$x=2,-5,7$

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, $x=2,-5,7$