Polynomial division : finding the unknown parameters in polynomial via divisibility (Division statement)

Question

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}

Answer 1

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 $m=0$, $1-p+q=0$.

Answer 2

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 $$m^2=2-p\\q=1$$

Answer 3

The hint:

We need $$x^4+px^2+q=(x^2+mx+1)(x^2-mx+q)$$

Answer 4

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 $x^2 + mx + 1 \mid x^4 + px^2 + q$.