Question 824812
Let {{{a}}} be the other zero of the polynomial {{{x^3+px+q}}} .
{{{(x-2)^2(x-a)=x^3+px+q}}}
{{{(x^2-4x+4)(x-a)=x^3+px+q}}}
{{{x(x^2-4x+4)-a(x^2-4x+4)=x^3+px+q}}}
{{{x^3-4x^2+4x-ax^2+4ax-4a=x^3+px+q}}}
{{{x^3-4x^2-ax^2+4x+4ax-4a=x^3+px+q}}}
{{{x^3-(4+a)x^2+(4+4a)x-4a=x^3+px+q}}}
Since the polynomials above have the same value for all values of {{{x}}} ,
the coefficients of all terms must be the same:
{{{1=1}}} for the term in {{{x^3}}} ,
{{{-(4+a)=0}}} for the term in {{{x^2}}} ,
{{{4+4a=p}}} for the term in {{{x}}} , and
{{{-4a=q}}} for the independent term.
 
Solving for {{{a}}} :
{{{-(4+a)=0}}} 
{{{4+a=0}}}
{{{highlight(a=-4)}}}
 
Then,
{{{p=4+4a}}}
{{{p=4+4(-4)}}}
{{{p=4-16}}}
{{{highlight(p=-12)}}}
and
{{{q=-4a}}}
{{{q=-4(-4)}}}
{{{highlight(q=16)}}}