Question 1157973
{{{P(x)=ax^3+bx^2+cx-4}}}
If you divide {{{P(x)}}} by {{{x+1}}} , you obtain a quotient {{{Q(x)}}} and remainder {{{r}}} that is a constant.
You may remember that in then {{{P(-1)=r}}}
If you did not remember, you would understand that if the reminder is {{{r}}} ,
it means that {{{P(x)=(x+1)Q(x)+r}}} and that
for {{{x=-1}}} , {{{P(-1)=(-1+1)Q(-1)+r=0*Q(-1)+r=0+r=r}}} .
So {{{a(-1)x^3+b(-1)^2+c(-1)-4=r}}}-->{{{highlight(-a+b-c-4=r)}}} .
Similarly the remainder, when dividing {{{P(x)}}} by {{{x+2}}} is
{{{a(-2)x^3+b(-2)^2+c(-2)-4=2r}}}-->{{{highlight(-8a+4b-2c-4=2r)}}} .
Then, {{{-8a+4b-2c-4=2(-a+b-c-4))}}}-->{{{-8a+4b-2c-4=-2a+2b-2c-8)}}}-->{{{ 4b-2b-2c+2c=8a-2a-8+4}}}-->{{{2b=6a-4}}}-->{{{highlight(b=3a-2)}}}