Two polynomials having a quadratic common divisor
$begingroup$
Find the real numbers a,b such that the polynomials
$$p(x)=x^4-2x^3+3x^2+2x+a
$$
$$q(x)=x^4-3x^3+4x^2+3x+b
$$
have a common divisor of degree two.
My attempt: Euclid algorithm:
we perform the divisions:
f:g obtain remainder r1 of degree 3
g:r1 obtain remainder r2 of degree 2
r1: r2 obtain remainder r3 of degree 1
And the condition must be that r3=0, but i get a not so nice system of equations with a and b
Is there any simpler method?
polynomials greatest-common-divisor
edited Jan 23 at 16:15
David Holden
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
$endgroup$
add a comment |
$begingroup$
Find the real numbers a,b such that the polynomials
$$p(x)=x^4-2x^3+3x^2+2x+a
$$
$$q(x)=x^4-3x^3+4x^2+3x+b
$$
have a common divisor of degree two.
My attempt: Euclid algorithm:
we perform the divisions:
f:g obtain remainder r1 of degree 3
g:r1 obtain remainder r2 of degree 2
r1: r2 obtain remainder r3 of degree 1
And the condition must be that r3=0, but i get a not so nice system of equations with a and b
Is there any simpler method?
polynomials greatest-common-divisor
edited Jan 23 at 16:15
David Holden
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
$endgroup$
$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21
add a comment |
$begingroup$
Find the real numbers a,b such that the polynomials
$$p(x)=x^4-2x^3+3x^2+2x+a
$$
$$q(x)=x^4-3x^3+4x^2+3x+b
$$
have a common divisor of degree two.
My attempt: Euclid algorithm:
we perform the divisions:
f:g obtain remainder r1 of degree 3
g:r1 obtain remainder r2 of degree 2
r1: r2 obtain remainder r3 of degree 1
And the condition must be that r3=0, but i get a not so nice system of equations with a and b
Is there any simpler method?
polynomials greatest-common-divisor
edited Jan 23 at 16:15
David Holden
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
$endgroup$
Find the real numbers a,b such that the polynomials
$$p(x)=x^4-2x^3+3x^2+2x+a
$$
$$q(x)=x^4-3x^3+4x^2+3x+b
$$
have a common divisor of degree two.
My attempt: Euclid algorithm:
we perform the divisions:
f:g obtain remainder r1 of degree 3
g:r1 obtain remainder r2 of degree 2
r1: r2 obtain remainder r3 of degree 1
And the condition must be that r3=0, but i get a not so nice system of equations with a and b
Is there any simpler method?
polynomials greatest-common-divisor
polynomials greatest-common-divisor
edited Jan 23 at 16:15
David Holden
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
edited Jan 23 at 16:15
David Holden
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
edited Jan 23 at 16:15
David Holden
14.9k21224
edited Jan 23 at 16:15
David Holden
14.9k21224
edited Jan 23 at 16:15
David Holden
14.9k21224
14.9k21224
asked Jan 23 at 16:12
amarius8312amarius8312
1827
asked Jan 23 at 16:12
amarius8312amarius8312
1827
asked Jan 23 at 16:12
amarius8312amarius8312
1827
1827
$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21
add a comment |
$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21
$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21
$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21
add a comment |
1 Answer
1
active
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votes
$begingroup$
That's the method that I would use. But, in fact, doing it that way I got the system:$$left{begin{array}{l}-2a^2+3ab+17a-b^2-13b=0\a^2-2ab-11a+b^2+8b=5.end{array}right.$$In order to solve it, I used Mathematica, which gave me $3$ solutions: $(a,b)=(-4,-5)$, $(a,b)=(5,-5)$, and $(a,b)=(5,7)$. And they are indeed solutions of the original problem.
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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votes
$begingroup$
That's the method that I would use. But, in fact, doing it that way I got the system:$$left{begin{array}{l}-2a^2+3ab+17a-b^2-13b=0\a^2-2ab-11a+b^2+8b=5.end{array}right.$$In order to solve it, I used Mathematica, which gave me $3$ solutions: $(a,b)=(-4,-5)$, $(a,b)=(5,-5)$, and $(a,b)=(5,7)$. And they are indeed solutions of the original problem.
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
That's the method that I would use. But, in fact, doing it that way I got the system:$$left{begin{array}{l}-2a^2+3ab+17a-b^2-13b=0\a^2-2ab-11a+b^2+8b=5.end{array}right.$$In order to solve it, I used Mathematica, which gave me $3$ solutions: $(a,b)=(-4,-5)$, $(a,b)=(5,-5)$, and $(a,b)=(5,7)$. And they are indeed solutions of the original problem.
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
That's the method that I would use. But, in fact, doing it that way I got the system:$$left{begin{array}{l}-2a^2+3ab+17a-b^2-13b=0\a^2-2ab-11a+b^2+8b=5.end{array}right.$$In order to solve it, I used Mathematica, which gave me $3$ solutions: $(a,b)=(-4,-5)$, $(a,b)=(5,-5)$, and $(a,b)=(5,7)$. And they are indeed solutions of the original problem.
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
That's the method that I would use. But, in fact, doing it that way I got the system:$$left{begin{array}{l}-2a^2+3ab+17a-b^2-13b=0\a^2-2ab-11a+b^2+8b=5.end{array}right.$$In order to solve it, I used Mathematica, which gave me $3$ solutions: $(a,b)=(-4,-5)$, $(a,b)=(5,-5)$, and $(a,b)=(5,7)$. And they are indeed solutions of the original problem.
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 16:32
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
add a comment |
add a comment |
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$begingroup$
using polynomial remainder theorem with x-(-d(x-1)*x-e*x-c) doesn't work ? oh and parity arguments can be used to show the first won't be even unless a is, and the second one alternates regardless of b's parity.
$endgroup$
– Roddy MacPhee
Feb 21 at 19:21