Question

The polynomials $a{x}^3+4x^2+3x-4$ and $x^3-4x+a$ leave the same remainder when divided by $(x-3)$. Find the value of $a$.

Answer 1

Given polynomials:
$p\left(x\right)=a{x}^3+4x^2+3x-4$
$q\left(x\right)=x^3-4x+a$
Using remainder theorem, the remainders when $p\left(x\right)$ and $q\left(x\right)$ are divided by $(x-3)$ are $p\left(3\right)$ and $q\left(3\right)$ respectively.
$p\left(3\right)=a\times (3{)}^3+4\times (3{)}^2+3\times 3-4$
$p\left(3\right)=27a+41$
$q\left(3\right)=(3{)}^3-4\times 3+a$
$q\left(3\right)=15+a$
Given that the remainder is the same.
$\Rightarrow p\left(3\right)=q\left(3\right)$
$\Rightarrow 27a+41=15+a$
$\Rightarrow 26a=-26$
$\Rightarrow a=-1$