Polynomial division : finding the unknown parameters in polynomial via divisibility (Division statement) 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}
 A: 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 $m=0$, $1-p+q=0$.
A: The hint:
We need $$x^4+px^2+q=(x^2+mx+1)(x^2-mx+q)$$
A: If the quartic polynomial $x^4 + px^2 + q$ is divisible by a quadratic of the form $x^2 + mx + 1$, so that
$x^4 + px^2 + q = f(x)(x^2 + mx + 1) \tag 1$
for some polynomial $f(x)$, severe restrictions are placed upon $x^4 + px^2 + q$, $x^2 + mx + 1$, and $f(x)$, as is seen below:
First of all, the relation (1) implies that $f(x)$ must itself be a monic quadratic; clearly we have
$\deg f(x) = 2, \tag 2$
and furthermore, writing
$f(x) = lx^2 + ax + b, \tag 3$
we see that the leading coefficient $l$ of $f(x)$, must obey
$l = l(1) = 1, \tag 4$
by comparing coeffiecients  $x^4$; therefore we may write
$f(x) = x^2 + ax + b, \tag 5$
and (1) becomes
$ x^4 + px^2 + q = (x^2 + mx + 1)(x^2 + ax + b)$
$= x^4 + ax^3 + bx^2 + mx^3 + am x^2 + bmx + x^2 + ax + b$
$= x^4 + (a + m)x^3 + (b + am + 1)x^2 + (bm + a)x + b; \tag 6$
so, comparing coeffiecients of powers of $x$ on either side we deduce that
$b = q, \tag 7$
$bm + a = 0, \tag 8$
$b + am + 1 = p, \tag 9$
$a + m = 0; \tag{10}$
from (10),
$a = -m; \tag{11}$
using this together with (7) in (9):
$q - m^2 + 1 = p, \tag{12}$
or
$m^2 = q - p + 1; \tag{13}$
from (8) and (11),
$m(b - 1) = mb - m = mb + a = 0; \tag{14}$
so, via (7),
$m(q - 1) = 0; \tag{15}$
we now branch according to the value of $q$; if 
$q \ne 1, \tag{16}$
then
$m = 0, \tag{17}$
so by (11),
$a = 0; \tag{18}$
and from (13),
$p = q + 1, \tag{19}$
thus again invoking (7) we find
$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); \tag{15}$
on the other hand, if
$q = 1, \tag{16}$
we have by (12) that
$p = 2 - m^2, \tag{17}$
so (9) and (11) lead us to
$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, \tag{18}$
and finally,
$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). \tag{19}$
(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 \mid x^4 + px^2 + q$.
A: By consecutive division we obtain $$x^4+px^2+q=(x^2+mx+1)(x^2-mx+m^2+p-1)+x(2m-mp-m^3)+q-p-m^2+1$$the remainder must be zero therefore $$q+1=m^2+p\\m(2-p)=m^3$$then either $$m=0\\p=q+1$$ or $$m^2=2-p\\q=1$$
