Question 513921
You really have a "quartic" (fourth power) equation here!
{{{x(x-2)^3-35(x-2)^2 = 0}}} First, factor the common factor {{{(x-2)^2}}}
{{{(x-2)^2(x(x-2)-35) = 0}}} Simplify.
{{{(x-2)^2(x^2-2x-35) = 0}}} Apply the "zero product" rule.
{{{(x-2)^2 = 0}}} or {{{(x^2-2x-35) = 0}}} so...
{{{(x-2)^2 = 0}}}
{{{(x-2)(x-2) = 0}}} and so {{{x = 2}}} {{{x = 2}}} a double root for this part!
{{{x^2-2x-35 = 0}}} Factor.
{{{(x+5)(x-7) = 0}}} Apply the "zero product" rule.
{{{x+5 = 0}}} or {{{x-7 = 0}}} so...
{{{x = -5}}} or {{{x = 7}}}
The four roots are:
{{{x = 2}}}, {{{x = 2}}}, {{{x = -5}}}, {{{x = 7}}}