Question

Question 1
If the polynomials $a{z}^3+4z^2+3z–4$ and $z^3–4z+a$ leave the same remainder when divided by z – 3, find the value of a.

Answer 1

Let $p_1\left(z\right)=a{z}^3+4z^2+3z–4$ and $p_2\left(z\right)=z^3–4z+a$
When we divide $p_1\left(z\right)byz–3$, then we get the remainder as $p_1\left(3\right)$.
Now, $p_1\left(3\right)=a(3{)}^3+4(3{)}^2+3\left(3\right)–4$
$=27a+36+9–4=27a+41$
When we divide $p_2\left(z\right)byz–3$, then we get the remainder as $p_2\left(3\right)$.
Now $p_2\left(3\right)=(3{)}^3–4(3)+a$
$=27–12+a=15+a$
According to the question, both the remainders are same.
i.e. $p_1\left(3\right)=p_2\left(3\right)$
$\therefore 27a+41=15+a$
$\Rightarrow 27a–a=15–41$
$\Rightarrow 26a=-26$
$\therefore a=-1$