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?
 A: You are on the right track. 
Let $P(z)$ represent your polynomial.
Then we have $$P(z)=(z+3)^2 Q(z)$$
Hence $P(-3)=0$.
Also, we've $$P'(z)=2(z+3)Q(z)+(z+3)^2Q'(z)$$
And hence, $P'(-3)=0$.
You have got two equations and two variables. Now you need to solve for $a$ and $b$.
A: Hint: Write $f(z)=(z^2 +6z+9)(z^2+cz+d)$, expand it and compare coefficients with the given polynomial. Then you obtain easy equations in $a,b,c,d$. And yes, the quartic polynomial has $z=-3$ as a double root. Plugging it in, it gives the linear equation $a = 5b - 5$. Now the above equations give $b=2$, hence $a=5$, and $c=1$, $d=5$.
A: use that $${z}^{2}+ \left( a+b-6 \right) z-2\,a-6\,b+27+{\frac {4\,a+28\,b-76}{z+
3}}+{\frac {6\,a-30\,b+30}{ \left( z+3 \right) ^{2}}}=\frac{z^4+(a+b)z^3+4az^2+(a+b+32)z+45}{z^2+6z+9}
$$
A: 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 $b=2$.
