Question

If the polynomials $a{z}^3+4z^2+3z-4$ and $z^3-4z+a$ leave the same remainder when divided by $z-3$, then find the value of $a$.

• A. $-1$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is A $-1$

Here, $p\left(z\right)=a{\left(z\right)}^3+4{\left(z\right)}^2+3z-4$, $q\left(z\right)={\left(z\right)}^3-4z+a$, and the zero of $z-3$ is $3$.
So, by the given condition
$p\left(3\right)=q\left(3\right)$
$a{\left(3\right)}^3+4{\left(3\right)}^2+3\left(3\right)-4={\left(3\right)}^3-4\left(3\right)+a$
$27a+4\times 9+9-4=27-12+a$
$27a+36+9-4=27-12+a$
$27a+45-4=15+a$
$27a+41=15+a$
$27a-a=15-41$
$26a=-26$
$a=\frac{-26}{26}$
$a=-1$