Here is a question that involves unknowns to be found out in a polynomial and in the divisor:
Find a relation between the constants $m$, $p$ and $q$, such that:
$x^4 + px^2 +q\space$ is divisible by $x^2+mx+1$.
This one I really don't know how approach, except maybe factoring the quadratic generally using the quadratic formula.
{Note that this problem is a pre-calculus problem, so one cannot make use of any Calculus}
Best Answer
Suppose $P(x)=Q(x)\cdot R(x)$, where $P,Q,R$ are polynomials of $x$.
Then as $Q(a)=0, P(a)=0$.
Now we know for some $a$, $$Q(a)=\color{red}{a^2+ma+1=0}\iff a^2=-(ma+1)\tag1$$
For such $a$ we have $$0=P(a)=(ma+1)^2-p(ma+1)+q$$ $$m^2a^2+m(2-p)a+(1-p+q)=0$$
For $m\neq0$, $$\color{red}{a^2+\frac{2-p}{m}a+\frac{1-p+q}{m^2}=0}\tag2$$
Now since $(1)$ and $(2)$ are quadratic equations with $a$ as the same roots, both equations must be the same and have the same coefficients.
Solving, we get
$${m^2=2-p}=1-p+q$$ $${q=1}$$
For $m=0$, $1-p+q=0$.