Question 100274
Before I start, if {{{ax^2+bx+3=0}}}, then {{{ax^2+bx=-3}}}. So keep this fact in mind



{{{p(b-ap)}}} Start with the left side of the second expression


={{{bp-ap^2}}} Distribute



Since we're assuming x+p is a factor, this means {{{x=-p}}} which means {{{p=-x}}}


={{{b(-x)-a(-x)^2}}} Plug in {{{p=-x}}}


={{{-bx-ax^2}}} Simplify



={{{-ax^2-bx}}} Rearrange the terms



={{{-(ax^2+bx)}}} Factor out a negative 1


Remember {{{ax^2+bx=-3}}}, so  let's substitute that in


={{{-(-3)}}} Replace {{{ax^2+bx}}} with -3


={{{3}}} Negate -3 to get 3


So we've algebraically manipulated {{{p(b-ap)}}} to get 3. So we've just shown that {{{p(b-ap)}}} is equal to 3



So that means we've proven that {{{p(b-ap)=3}}} is true.