Question

If a polynomial $a{z}^3+4z^2+2z-4$ and $z^2+4z+a$ leave the same remainder divide by $(z-3)$. Find the value of a.

• A. $1$
• B. $\frac{-17}{26}$
• C. $-1$
• D. $17$

Answer 1

Answer: (B)

$\frac{-17}{26}$

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

It is given that the zero of the polynomial $f\left(z\right)$ is $(z-3)$ that is $z=3$

So,replacing $z$ by $3$, we get

$f\left(3\right)=a{\left(3\right)}^3+4{\left(3\right)}^2+2\left(3\right)-4=27a+(4\times 9)+6-4=27a+36+6-4=27a+42-4=27a+38$

Let $g\left(z\right)=z^2+4z+a$

It is given that the zero of the polynomial $f\left(z\right)$ is $(z-3)$ that is $z=3$

So,replacing $z$ by $3$, we get

$g\left(3\right)={\left(3\right)}^2+(4\times 3)+a=9+12+a=21+a$

Now, since the given two polynomials leaves same remainder when divided by $z-3$.

Therefore, $g\left(3\right)=f\left(3\right)$ that is:

$27a+38=21+a\Rightarrow 27a-a=21-38\Rightarrow 26a=-17\Rightarrow a=-\frac{17}{26}$

Hence, $a=-\frac{17}{26}$.