Prove that p(x) cannot have more than m zeros , where $p(x)$ is a polynominal of degree m and $p(x) ∈...
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This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
The question is as stated in the title:
Prove that p(x) cannot have more than m zeros $p(x)$, where $p(x)$ is a polynominal of degree m and $p(x) ∈ mathbb{K} [x]$ and $mathbb{K}$ is a field . (where the zeros are counted with multiplcities)
I am also given the hint : Use the fact that $mathbb{K} [x]$ is a factorial ring
abstract-algebra
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
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marked as duplicate by Bill Dubuque abstract-algebra
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Jan 14 at 16:59
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.
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$begingroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
The question is as stated in the title:
Prove that p(x) cannot have more than m zeros $p(x)$, where $p(x)$ is a polynominal of degree m and $p(x) ∈ mathbb{K} [x]$ and $mathbb{K}$ is a field . (where the zeros are counted with multiplcities)
I am also given the hint : Use the fact that $mathbb{K} [x]$ is a factorial ring
abstract-algebra
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
$endgroup$
marked as duplicate by Bill Dubuque abstract-algebra
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Jan 14 at 16:59
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.
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Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
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– Aniruddh Agarwal
Jan 14 at 16:58
add a comment |
$begingroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
The question is as stated in the title:
Prove that p(x) cannot have more than m zeros $p(x)$, where $p(x)$ is a polynominal of degree m and $p(x) ∈ mathbb{K} [x]$ and $mathbb{K}$ is a field . (where the zeros are counted with multiplcities)
I am also given the hint : Use the fact that $mathbb{K} [x]$ is a factorial ring
abstract-algebra
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
$endgroup$
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
The question is as stated in the title:
Prove that p(x) cannot have more than m zeros $p(x)$, where $p(x)$ is a polynominal of degree m and $p(x) ∈ mathbb{K} [x]$ and $mathbb{K}$ is a field . (where the zeros are counted with multiplcities)
I am also given the hint : Use the fact that $mathbb{K} [x]$ is a factorial ring
This question already has an answer here:
Show that if $ R $ is an integral domain then a polynomial in $ R[X] $ of degree $ d $ can have at most $ d $ roots
2 answers
abstract-algebra
abstract-algebra
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
edited Jan 14 at 17:38
Aniruddh Agarwal
1218
1218
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
asked Jan 14 at 16:53
Christian LejdströmChristian Lejdström
1
1
marked as duplicate by Bill Dubuque abstract-algebra
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Jan 14 at 16:59
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 Bill Dubuque abstract-algebra
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Jan 14 at 16:59
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.
1
$begingroup$
Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
$endgroup$
– Aniruddh Agarwal
Jan 14 at 16:58
add a comment |
1
$begingroup$
Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
$endgroup$
– Aniruddh Agarwal
Jan 14 at 16:58
1
1
$begingroup$
Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
$endgroup$
– Aniruddh Agarwal
Jan 14 at 16:58
$begingroup$
Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
$endgroup$
– Aniruddh Agarwal
Jan 14 at 16:58
add a comment |
1 Answer
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Hint: Any root of a polynomial is intricately linked to a linear factor, and polynomials have degrees.
answered Jan 14 at 16:58
ArthurArthur
123k7122211
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Any root of a polynomial is intricately linked to a linear factor, and polynomials have degrees.
answered Jan 14 at 16:58
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
Hint: Any root of a polynomial is intricately linked to a linear factor, and polynomials have degrees.
answered Jan 14 at 16:58
ArthurArthur
123k7122211
$endgroup$
add a comment |
$begingroup$
Hint: Any root of a polynomial is intricately linked to a linear factor, and polynomials have degrees.
answered Jan 14 at 16:58
ArthurArthur
123k7122211
$endgroup$
Hint: Any root of a polynomial is intricately linked to a linear factor, and polynomials have degrees.
answered Jan 14 at 16:58
ArthurArthur
123k7122211
answered Jan 14 at 16:58
ArthurArthur
123k7122211
answered Jan 14 at 16:58
ArthurArthur
123k7122211
answered Jan 14 at 16:58
ArthurArthur
123k7122211
123k7122211
add a comment |
add a comment |
1
$begingroup$
Note that if $alpha$ is a root for $p(x)$, then $(x - alpha) mid p(x)$.
$endgroup$
– Aniruddh Agarwal
Jan 14 at 16:58