Question

If the polynomial $a{z}^3+4z^2+3z-4\text{and}{z}^3-4z+a$ leaves same remainder when divided by $z-3$ then find ${}^{\prime }{a}^{\prime }$.

Answer 1

$(z-3)$ leaves same remainder for two polynomials.

According to remainder theorem,

$z=3$ when substituted in the polynomials will give remainder.

Substituting $z=3$ in both polynomials.

$=a(3{)}^2+4(3{)}^2+3\left(3\right)-4$

$=27a+36+9-4=27a+41--------\left(1\right)$

Now in other polynomial

$=(3{)}^2-4(3)+a=27-12+a=15+a--------(2)$

According to question, both are equal

$⟹27a+41=15+a$

$⟹26a=-26$

$⟹a=-1$