Question 880364
A 3rd degree polynomial with zeros at {{{x=-2}}} , {{{x=4}}} , and {{{x=6}}} ,
must have the factoring {{{p(x)=a(x-(-2))(x-4)(x-6)=a(x+2)(x-4)(x-6)}}} .
If it "passes though the point (0,3)", then
{{{p(0)=3}}} , so
{{{a(0+2)(x-4)(x-6)=3}}}--->{{{2(-4)-6)a=3}}}--->{{{48a=3}}}--->{{{a=3/48}}}-->{{{a=1/6}}} .
So the polynomial is
{{{p(x)=(1/16)(x+2)(x-4)(x-6)}}} .