<s>[INST] If polynomial is divisible by quadratic, find values of $a$ and $b$ Equation is $z^4+(a+b)z^3+4az^2+(a+b+32)z+45$ which is divisible by $z^2+6z+9$.
We're meant to find values of $a$ and $b$ and then solve the equation itself!
I figured $z^2+6z+9=(z+3)^2$, but was wondering if this automatically implies that it is a double root(?) If so, would differentiating the equation and do the normal simultaneous equations thing suitable for this question?
 [/INST] Let $p(z)=z^4+(a+b)z^3+4az^2+(a+b+32)z+45$. Then\begin{multline}p(z)=p\bigl((z+3)-3\bigr)=(z+3)^4+(a+b-12)(z+3)^3+\\+(-5a-9b+54)(z+3)^2+(4a+28b-76)(z+3)+6a-30b+30.\end{multline}On the other hand, $z^2+6z+9=(z+3)^2$. Therefore, $p(z)$ is a multiple of $z^2+6z+9$ if and only if$$\left\{\begin{array}{l}4a+28b-76=0\\6a-30b+30=0.\end{array}\right.$$This means that $a=5$ and that $b=2$.
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