Mixing
Silly question: are polynomials “X” and “X^2” reducible?
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Since f(0)=0 holds for both, are they both reducible polynomials?
I'm asking because I'm working on this question where they're asking you to factor the following in irreducible factors.
X^5 - X^4 - X^3 + X^2 + X ∈ F3[X]
The solution is X(X^2 + X + 2)^2, so that includes factor X, which I believe is reducible... Am I missing something here?
polynomials irreducible-polynomials
asked Jan 20 at 15:31
SJ19SJ19
1224
$endgroup$
add a comment |
$begingroup$
Since f(0)=0 holds for both, are they both reducible polynomials?
I'm asking because I'm working on this question where they're asking you to factor the following in irreducible factors.
X^5 - X^4 - X^3 + X^2 + X ∈ F3[X]
The solution is X(X^2 + X + 2)^2, so that includes factor X, which I believe is reducible... Am I missing something here?
polynomials irreducible-polynomials
asked Jan 20 at 15:31
SJ19SJ19
1224
$endgroup$
2
$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38
add a comment |
$begingroup$
Since f(0)=0 holds for both, are they both reducible polynomials?
I'm asking because I'm working on this question where they're asking you to factor the following in irreducible factors.
X^5 - X^4 - X^3 + X^2 + X ∈ F3[X]
The solution is X(X^2 + X + 2)^2, so that includes factor X, which I believe is reducible... Am I missing something here?
polynomials irreducible-polynomials
asked Jan 20 at 15:31
SJ19SJ19
1224
$endgroup$
Since f(0)=0 holds for both, are they both reducible polynomials?
I'm asking because I'm working on this question where they're asking you to factor the following in irreducible factors.
X^5 - X^4 - X^3 + X^2 + X ∈ F3[X]
The solution is X(X^2 + X + 2)^2, so that includes factor X, which I believe is reducible... Am I missing something here?
polynomials irreducible-polynomials
polynomials irreducible-polynomials
asked Jan 20 at 15:31
SJ19SJ19
1224
asked Jan 20 at 15:31
SJ19SJ19
1224
asked Jan 20 at 15:31
SJ19SJ19
1224
asked Jan 20 at 15:31
SJ19SJ19
1224
asked Jan 20 at 15:31
SJ19SJ19
1224
1224
2
$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38
add a comment |
2
$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38
2
2
$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38
add a comment |
1 Answer
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$begingroup$
As any other polynomial of degree one (over some field) , $x$ is irreducible, whereas $x^2=xcdot x$ is reducible
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
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add a comment |
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$begingroup$
As any other polynomial of degree one (over some field) , $x$ is irreducible, whereas $x^2=xcdot x$ is reducible
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
As any other polynomial of degree one (over some field) , $x$ is irreducible, whereas $x^2=xcdot x$ is reducible
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
As any other polynomial of degree one (over some field) , $x$ is irreducible, whereas $x^2=xcdot x$ is reducible
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
$endgroup$
As any other polynomial of degree one (over some field) , $x$ is irreducible, whereas $x^2=xcdot x$ is reducible
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
answered Jan 20 at 16:06
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
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$begingroup$
As any other polynomial of degree one (over some field) , $;x;$ is irreducible, whereas $;x^2=xcdot x;$ is reducible
$endgroup$
– DonAntonio
Jan 20 at 15:33
$begingroup$
Thank you. Guess I got blinded by all the rules I studied and kinda forgot about the simple ones. If you put that in an answer I'll approve it.
$endgroup$
– SJ19
Jan 20 at 15:38