Question

The polynomials $p\left(x\right)=4x^3-2x^2+px+5and,q\left(x\right)=x^3+6x^2+p$ leave the remainders $a$ and $b$ respectively, when divided by $(x+2)$. Find the value of $p$ if $a+b=0$.

Answer 1

Remainder theorem:

When we divide $f\left(x\right)$ by $(x-c)$ , the remainder is equal to $f\left(c\right)$

When $p\left(x\right)=4x^3-2x^2+px+5$ is divided by $(x+2)$, the remainder is

$p(-2)=4(-2{)}^3-2(-2{)}^2+p(-2)+5$

$=-2p-35$

So,

$a=-2p-35$

When $q\left(x\right)=x^3+6x^2+p$ is divided by $(x+2)$, the remainder is

$q(-2)=(-2{)}^3+6(-2{)}^2+p$

$=p+16$

So,

$b=p+16$

Now,

$a+b=0$

$-2p-35+p+16=0$

$-p-19=0$

$p=-19$