Question

For which values of $-a$ are the zeroes of $q\left(x\right)=x^3+2x^2+a$ also the zeroes of the polynomial $p\left(x\right)=x^5-x^4-4x^3+3x^2+3x+b?$

Answer 1

Given that the zeroes of $q\left(x\right)=x^3+2x^2+a$ are also the zeroes of the polynomial $p\left(x\right)=x^5–x^4–4x^3+3x^2+3x+b$ that is $q\left(x\right)$ is a factor of $p\left(x\right)$. Then, we use a division algorithm as follows:

If $(x^3+2x^2+a)$ is a factor of $(x^5–x^4–4x^3+3x^2+3x+b)$, then remainder should be zero that is:

$–(1+a)x^2+(3+3a)x+(b–2a)=0$

Comparing the coefficient of $x$ to find the value of $a$, we have:

$3+3a=0\Rightarrow 3a=-3\Rightarrow a=\frac{-3}{3}\Rightarrow a=-1$

Hence, $a=-1$.