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

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

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:

$x^4 + px^2 +qspace$ 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

edited 2 days ago

asked 2 days ago

Palu

2892519

add a commentÂ |Â

up vote
1
down vote

favorite

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:

$x^4 + px^2 +qspace$ 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

edited 2 days ago

asked 2 days ago

Palu

2892519

add a commentÂ |Â

up vote
1
down vote

favorite

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:

$x^4 + px^2 +qspace$ 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

edited 2 days ago

asked 2 days ago

Palu

2892519

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:

$x^4 + px^2 +qspace$ 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

edited 2 days ago

asked 2 days ago

Palu

2892519

edited 2 days ago

asked 2 days ago

Palu

2892519

asked 2 days ago

Palu

2892519

asked 2 days ago

Palu

2892519

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
2
down vote

accepted

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 $a$,
$$Q(a)=colorreda^2+ma+1=0iff a^2=-(ma+1)tag1$$

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 $mneq0$,
$$colorreda^2+frac2-pma+frac1-p+qm^2=0tag2$$

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

answered yesterday

Mythomorphic

5,0361532

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

add a commentÂ |Â

up vote
1
down vote

The hint:

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

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

add a commentÂ |Â

up vote
1
down vote

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

$b = q, tag 7$

$bm + a = 0, tag 8$

$b + am + 1 = p, tag 9$

$a + m = 0; tag10$

from (10),

$a = -m; tag11$

using this together with (7) in (9):

$q - m^2 + 1 = p, tag12$

$m^2 = q - p + 1; tag13$

from (8) and (11),

$m(b - 1) = mb - m = mb + a = 0; tag14$

so, via (7),

$m(q - 1) = 0; tag15$

we now branch according to the value of $q$; if

$q ne 1, tag16$

then

$m = 0, tag17$

so by (11),

$a = 0; tag18$

and from (13),

$p = q + 1, tag19$

thus again invoking (7) we find

$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); tag15$

on the other hand, if

$q = 1, tag16$

we have by (12) that

$p = 2 - m^2, tag17$

so (9) and (11) lead us to

$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, tag18$

and finally,

$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). tag19$

(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 mid x^4 + px^2 + q$.

answered yesterday

Robert Lewis

36.7k22155

add a commentÂ |Â

up vote
0
down vote

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

answered 2 days ago

Mostafa Ayaz

8,5203530

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

add a commentÂ |Â

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872219%2fpolynomial-division-finding-the-unknown-parameters-in-polynomial-via-divisibil%23new-answer', 'question_page');

);

Post as a guest

Name

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
2
down vote

accepted

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 $a$,
$$Q(a)=colorreda^2+ma+1=0iff a^2=-(ma+1)tag1$$

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 $mneq0$,
$$colorreda^2+frac2-pma+frac1-p+qm^2=0tag2$$

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

answered yesterday

Mythomorphic

5,0361532

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

add a commentÂ |Â

up vote
2
down vote

accepted

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 $a$,
$$Q(a)=colorreda^2+ma+1=0iff a^2=-(ma+1)tag1$$

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 $mneq0$,
$$colorreda^2+frac2-pma+frac1-p+qm^2=0tag2$$

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

answered yesterday

Mythomorphic

5,0361532

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

add a commentÂ |Â

up vote
2
down vote

accepted

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 $a$,
$$Q(a)=colorreda^2+ma+1=0iff a^2=-(ma+1)tag1$$

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 $mneq0$,
$$colorreda^2+frac2-pma+frac1-p+qm^2=0tag2$$

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

answered yesterday

Mythomorphic

5,0361532

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 $a$,
$$Q(a)=colorreda^2+ma+1=0iff a^2=-(ma+1)tag1$$

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 $mneq0$,
$$colorreda^2+frac2-pma+frac1-p+qm^2=0tag2$$

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

answered yesterday

Mythomorphic

5,0361532

answered yesterday

Mythomorphic

5,0361532

answered yesterday

Mythomorphic

5,0361532

answered yesterday

Mythomorphic

5,0361532

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

add a commentÂ |Â

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

Hi Mythomorphic, this is a very interesting approach, your algebraic manipulations leading to another quadratic is very very interesting! Very Beautiful approach in my opinion.
â€“Â Palu
yesterday

If you have learnt Remainder Theorem, you should find it familiar:)
â€“Â Mythomorphic
yesterday

Yes. And remainder theorem along with factor theorem are the tools one can use on this problem.
â€“Â Palu
yesterday

add a commentÂ |Â

up vote
1
down vote

The hint:

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

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

add a commentÂ |Â

up vote
1
down vote

The hint:

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

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

add a commentÂ |Â

up vote
1
down vote

The hint:

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

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

The hint:

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

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

edited 2 days ago

answered 2 days ago

Michael Rozenberg

86.9k1575178

answered 2 days ago

Michael Rozenberg

86.9k1575178

answered 2 days ago

Michael Rozenberg

86.9k1575178

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

add a commentÂ |Â

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

Hi Michael, yes this is very good. I did not think of this kind of factorization.
â€“Â Palu
2 days ago

add a commentÂ |Â

up vote
1
down vote

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

$b = q, tag 7$

$bm + a = 0, tag 8$

$b + am + 1 = p, tag 9$

$a + m = 0; tag10$

from (10),

$a = -m; tag11$

using this together with (7) in (9):

$q - m^2 + 1 = p, tag12$

$m^2 = q - p + 1; tag13$

from (8) and (11),

$m(b - 1) = mb - m = mb + a = 0; tag14$

so, via (7),

$m(q - 1) = 0; tag15$

we now branch according to the value of $q$; if

$q ne 1, tag16$

then

$m = 0, tag17$

so by (11),

$a = 0; tag18$

and from (13),

$p = q + 1, tag19$

thus again invoking (7) we find

$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); tag15$

on the other hand, if

$q = 1, tag16$

we have by (12) that

$p = 2 - m^2, tag17$

so (9) and (11) lead us to

$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, tag18$

and finally,

$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). tag19$

(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 mid x^4 + px^2 + q$.

answered yesterday

Robert Lewis

36.7k22155

add a commentÂ |Â

up vote
1
down vote

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

$b = q, tag 7$

$bm + a = 0, tag 8$

$b + am + 1 = p, tag 9$

$a + m = 0; tag10$

from (10),

$a = -m; tag11$

using this together with (7) in (9):

$q - m^2 + 1 = p, tag12$

$m^2 = q - p + 1; tag13$

from (8) and (11),

$m(b - 1) = mb - m = mb + a = 0; tag14$

so, via (7),

$m(q - 1) = 0; tag15$

we now branch according to the value of $q$; if

$q ne 1, tag16$

then

$m = 0, tag17$

so by (11),

$a = 0; tag18$

and from (13),

$p = q + 1, tag19$

thus again invoking (7) we find

$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); tag15$

on the other hand, if

$q = 1, tag16$

we have by (12) that

$p = 2 - m^2, tag17$

so (9) and (11) lead us to

$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, tag18$

and finally,

$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). tag19$

(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 mid x^4 + px^2 + q$.

answered yesterday

Robert Lewis

36.7k22155

add a commentÂ |Â

up vote
1
down vote

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

$b = q, tag 7$

$bm + a = 0, tag 8$

$b + am + 1 = p, tag 9$

$a + m = 0; tag10$

from (10),

$a = -m; tag11$

using this together with (7) in (9):

$q - m^2 + 1 = p, tag12$

$m^2 = q - p + 1; tag13$

from (8) and (11),

$m(b - 1) = mb - m = mb + a = 0; tag14$

so, via (7),

$m(q - 1) = 0; tag15$

we now branch according to the value of $q$; if

$q ne 1, tag16$

then

$m = 0, tag17$

so by (11),

$a = 0; tag18$

and from (13),

$p = q + 1, tag19$

thus again invoking (7) we find

$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); tag15$

on the other hand, if

$q = 1, tag16$

we have by (12) that

$p = 2 - m^2, tag17$

so (9) and (11) lead us to

$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, tag18$

and finally,

$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). tag19$

(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 mid x^4 + px^2 + q$.

answered yesterday

Robert Lewis

36.7k22155

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

$b = q, tag 7$

$bm + a = 0, tag 8$

$b + am + 1 = p, tag 9$

$a + m = 0; tag10$

from (10),

$a = -m; tag11$

using this together with (7) in (9):

$q - m^2 + 1 = p, tag12$

$m^2 = q - p + 1; tag13$

from (8) and (11),

$m(b - 1) = mb - m = mb + a = 0; tag14$

so, via (7),

$m(q - 1) = 0; tag15$

we now branch according to the value of $q$; if

$q ne 1, tag16$

then

$m = 0, tag17$

so by (11),

$a = 0; tag18$

and from (13),

$p = q + 1, tag19$

thus again invoking (7) we find

$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); tag15$

on the other hand, if

$q = 1, tag16$

we have by (12) that

$p = 2 - m^2, tag17$

so (9) and (11) lead us to

$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, tag18$

and finally,

$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). tag19$

(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 mid x^4 + px^2 + q$.

answered yesterday

Robert Lewis

36.7k22155

answered yesterday

Robert Lewis

36.7k22155

answered yesterday

Robert Lewis

36.7k22155

answered yesterday

Robert Lewis

36.7k22155

add a commentÂ |Â

up vote
0
down vote

answered 2 days ago

Mostafa Ayaz

8,5203530

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

add a commentÂ |Â

up vote
0
down vote

answered 2 days ago

Mostafa Ayaz

8,5203530

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

add a commentÂ |Â

up vote
0
down vote

answered 2 days ago

Mostafa Ayaz

8,5203530

answered 2 days ago

Mostafa Ayaz

8,5203530

answered 2 days ago

Mostafa Ayaz

8,5203530

answered 2 days ago

Mostafa Ayaz

8,5203530

answered 2 days ago

Mostafa Ayaz

8,5203530

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

add a commentÂ |Â

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

I am very impressed with all the answers here!!! I wish i could pick many of these as the solution to the problem I have posed. But i think I will go with Mythomorphic solution, because it seems to be unique.
â€“Â Palu
yesterday

Alright but notice that also Mythomorphic's solution isn't unique. It leads to two...
â€“Â Mostafa Ayaz
yesterday

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

ukmuiik