Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $mathbb{Q}$? [duplicate]
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This question already has an answer here:
Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
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Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $mathbb{Q}$?
Can I say that there does not exist a prime that divides both $2$ and $1$? Or is there another way to show it is reducible? Do I just use rational root test?
field-theory irreducible-polynomials
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
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marked as duplicate by Did, amWhy, Martin Sleziak, Brahadeesh, user10354138 Dec 2 '18 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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This question already has an answer here:
Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
1 answer
Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $mathbb{Q}$?
Can I say that there does not exist a prime that divides both $2$ and $1$? Or is there another way to show it is reducible? Do I just use rational root test?
field-theory irreducible-polynomials
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
$endgroup$
marked as duplicate by Did, amWhy, Martin Sleziak, Brahadeesh, user10354138 Dec 2 '18 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
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Eisenstein's criterion is for irreducibility, not reducibility.
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– Lord Shark the Unknown
Dec 1 '18 at 20:12
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And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
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– Arthur
Dec 1 '18 at 20:24
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See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
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– Martin Sleziak
Dec 2 '18 at 4:05
add a comment |
$begingroup$
This question already has an answer here:
Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
1 answer
Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $mathbb{Q}$?
Can I say that there does not exist a prime that divides both $2$ and $1$? Or is there another way to show it is reducible? Do I just use rational root test?
field-theory irreducible-polynomials
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
$endgroup$
This question already has an answer here:
Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
1 answer
Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $mathbb{Q}$?
Can I say that there does not exist a prime that divides both $2$ and $1$? Or is there another way to show it is reducible? Do I just use rational root test?
This question already has an answer here:
Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
1 answer
field-theory irreducible-polynomials
field-theory irreducible-polynomials
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
asked Dec 1 '18 at 20:06
numericalorangenumericalorange
1,765311
1,765311
marked as duplicate by Did, amWhy, Martin Sleziak, Brahadeesh, user10354138 Dec 2 '18 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did, amWhy, Martin Sleziak, Brahadeesh, user10354138 Dec 2 '18 at 6:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
$begingroup$
Eisenstein's criterion is for irreducibility, not reducibility.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 20:12
$begingroup$
And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
$endgroup$
– Arthur
Dec 1 '18 at 20:24
$begingroup$
See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 4:05
add a comment |
5
$begingroup$
Eisenstein's criterion is for irreducibility, not reducibility.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 20:12
$begingroup$
And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
$endgroup$
– Arthur
Dec 1 '18 at 20:24
$begingroup$
See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 4:05
5
5
$begingroup$
Eisenstein's criterion is for irreducibility, not reducibility.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 20:12
$begingroup$
Eisenstein's criterion is for irreducibility, not reducibility.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 20:12
$begingroup$
And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
$endgroup$
– Arthur
Dec 1 '18 at 20:24
$begingroup$
And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
$endgroup$
– Arthur
Dec 1 '18 at 20:24
$begingroup$
See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 4:05
$begingroup$
See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 4:05
add a comment |
1 Answer
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You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate:
$$ (x^2 + ax + 1)(x^2 + bx + 4),$$
$$ (x^2 + ax + 2)(x^2 + bx + 2),$$
$$ (x^2 + ax - 1)(x^2 + bx - 4),$$
$$ (x^2 + ax - 2)(x^2 + bx -2),$$
where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
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1
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Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
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– numericalorange
Dec 1 '18 at 21:07
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@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
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– Will Jagy
Dec 1 '18 at 21:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate:
$$ (x^2 + ax + 1)(x^2 + bx + 4),$$
$$ (x^2 + ax + 2)(x^2 + bx + 2),$$
$$ (x^2 + ax - 1)(x^2 + bx - 4),$$
$$ (x^2 + ax - 2)(x^2 + bx -2),$$
where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
$endgroup$
1
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
add a comment |
$begingroup$
You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate:
$$ (x^2 + ax + 1)(x^2 + bx + 4),$$
$$ (x^2 + ax + 2)(x^2 + bx + 2),$$
$$ (x^2 + ax - 1)(x^2 + bx - 4),$$
$$ (x^2 + ax - 2)(x^2 + bx -2),$$
where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
$endgroup$
1
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
add a comment |
$begingroup$
You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate:
$$ (x^2 + ax + 1)(x^2 + bx + 4),$$
$$ (x^2 + ax + 2)(x^2 + bx + 2),$$
$$ (x^2 + ax - 1)(x^2 + bx - 4),$$
$$ (x^2 + ax - 2)(x^2 + bx -2),$$
where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
$endgroup$
You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
The second result states that if a non-constant polynomial with
integer coefficients is irreducible over the integers, then it is also
irreducible if it is considered as a polynomial over the rationals.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate:
$$ (x^2 + ax + 1)(x^2 + bx + 4),$$
$$ (x^2 + ax + 2)(x^2 + bx + 2),$$
$$ (x^2 + ax - 1)(x^2 + bx - 4),$$
$$ (x^2 + ax - 2)(x^2 + bx -2),$$
where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
edited Dec 1 '18 at 21:02
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
answered Dec 1 '18 at 20:49
Will JagyWill Jagy
103k5102200
103k5102200
1
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
add a comment |
1
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
1
1
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
Thanks for taking the time to reply and help me! I understand now and excuse me for my elementary knowledge on the subject!
$endgroup$
– numericalorange
Dec 1 '18 at 21:07
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
$begingroup$
@numericalorange please let me know when you finish it, and what you come to conclude about factoring.
$endgroup$
– Will Jagy
Dec 1 '18 at 21:16
add a comment |
5
$begingroup$
Eisenstein's criterion is for irreducibility, not reducibility.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 20:12
$begingroup$
And rational root test only gives you (or excludes the existence of) linear factors. This polynomial may be the product of two irreducible quadratic factors, and the rational root test doesn't say (as easily) anything about that.
$endgroup$
– Arthur
Dec 1 '18 at 20:24
$begingroup$
See also: Factoring/Reducing a polynomial $x^4 -2x^3 + 2x^2 + x + 4$
$endgroup$
– Martin Sleziak
Dec 2 '18 at 4:05