If the polynomials $az_{3}+4z_{2}+3z−4$ 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

$1$

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Solution

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Correct option is

A

$−1$

Here, $p(z)=a(z)_{3}+4(z)_{2}+3z−4$, $q(z)=(z)_{3}−4z+a$, and the zero of $z−3$ is $3$. So, by the given condition $p(3)=q(3)$ $a(3)_{3}+4(3)_{2}+3(3)−4=(3)_{3}−4(3)+a$ $27a+4×9+9−4=27−12+a$ $27a+36+9−4=27−12+a$ $27a+45−4=15+a$ $27a+41=15+a$ $27a−a=15−41$ $26a=−26$ $a=26−26 $ $a=−1$