$f(x)=a_nx^n+a_{n-1}x^{n-1}+…+a_1x+a_0$ and $g(x)=a_0x^n+a_1x^{n-1}+…+a_{n-1}x^{n-1}+a_n$ [closed]

-2

Suppose $f,g$ are two polynomial, given by $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and $g(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x^{n-1}+a_n$ where $a_0,a_1,...,a_n$ are elements of a field $F$.

Is it true that both $f$ and $g$ have same roots?

abstract-algebra polynomials

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asked Nov 21 '18 at 19:33

JhonJhon

22

closed as off-topic by Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R Nov 22 '18 at 2:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried simple examples?
  – WhatToDo
  Nov 21 '18 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint $ g(x) = x^n f(1/x) $
  – Bill Dubuque
  Nov 21 '18 at 21:31
  
  
  

  
  
  
  

add a comment | 

-2

Suppose $f,g$ are two polynomial, given by $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and $g(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x^{n-1}+a_n$ where $a_0,a_1,...,a_n$ are elements of a field $F$.

Is it true that both $f$ and $g$ have same roots?

abstract-algebra polynomials

share|cite|improve this question

asked Nov 21 '18 at 19:33

JhonJhon

22

closed as off-topic by Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R Nov 22 '18 at 2:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried simple examples?
  – WhatToDo
  Nov 21 '18 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint $ g(x) = x^n f(1/x) $
  – Bill Dubuque
  Nov 21 '18 at 21:31
  
  
  

  
  
  
  

add a comment | 

-2

-2

-2

Suppose $f,g$ are two polynomial, given by $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and $g(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x^{n-1}+a_n$ where $a_0,a_1,...,a_n$ are elements of a field $F$.

Is it true that both $f$ and $g$ have same roots?

abstract-algebra polynomials

share|cite|improve this question

asked Nov 21 '18 at 19:33

JhonJhon

22

Suppose $f,g$ are two polynomial, given by $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and $g(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x^{n-1}+a_n$ where $a_0,a_1,...,a_n$ are elements of a field $F$.

Is it true that both $f$ and $g$ have same roots?

abstract-algebra polynomials

abstract-algebra polynomials

share|cite|improve this question

asked Nov 21 '18 at 19:33

JhonJhon

22

share|cite|improve this question

asked Nov 21 '18 at 19:33

JhonJhon

22

share|cite|improve this question

share|cite|improve this question

asked Nov 21 '18 at 19:33

JhonJhon

22

asked Nov 21 '18 at 19:33

JhonJhon

22

asked Nov 21 '18 at 19:33

JhonJhon

22

22

closed as off-topic by Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R Nov 22 '18 at 2:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R Nov 22 '18 at 2:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, amWhy, Leucippus, KReiser, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried simple examples?
  – WhatToDo
  Nov 21 '18 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint $ g(x) = x^n f(1/x) $
  – Bill Dubuque
  Nov 21 '18 at 21:31
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried simple examples?
  – WhatToDo
  Nov 21 '18 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint $ g(x) = x^n f(1/x) $
  – Bill Dubuque
  Nov 21 '18 at 21:31
  
  
  

  
  
  
  

2

2

Have you tried simple examples?
– WhatToDo
Nov 21 '18 at 19:35

Have you tried simple examples?
– WhatToDo
Nov 21 '18 at 19:35

Hint $ g(x) = x^n f(1/x) $
– Bill Dubuque
Nov 21 '18 at 21:31

Hint $ g(x) = x^n f(1/x) $
– Bill Dubuque
Nov 21 '18 at 21:31

add a comment | 

3 Answers
3

active

oldest

votes

1

@Nicholas Stull

The quadratic equations $y=x^2+3x+2$ and $y=2x^2+3x+1$ both have the root $-1$. In fact, any quadratic equation $y=ax^2+bx+c$ for which $a+b=c$ will share one root if $aneq c$ (and a double root if $a=c$).

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
  – Nicholas Stull
  Nov 21 '18 at 21:14
  
  
  

  
  
  
  

add a comment | 

1

No. Consider the following cases over the field $F = mathbb{C}$

$f(x) = 3x^2 + 2x + 1$ and $g(x) = x^2 + 2x + 3$ do not have the same roots.

Even simpler, $f(x) = 2x+1$ and $g(x) = x+2$ do not share a root.

---

A (slightly) more interesting question:

If ${a_0,a_1,ldots,a_n}subset F$ are all distinct elements of the field $F$, do the polynomials $f(x) = a_0 x^n + ldots + a_n$ and $g(x) = a_n x^n + ldots + a_0$ ever share a single root in the field $F$?

(Distinct is required here, because otherwise you could simply choose $a_0 = a_n$, $a_1 = a_2 = cdots = a_{n-1}$ and come up with a simple example of this. Apparently this is quite simple, for quadratics, as demonstrated above...)

share|cite|improve this answer

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

add a comment | 

0

What is true is that $x mapsto 1/x$ is a bijection between the nonzero roots of $f$ and $g$.

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

@Nicholas Stull

The quadratic equations $y=x^2+3x+2$ and $y=2x^2+3x+1$ both have the root $-1$. In fact, any quadratic equation $y=ax^2+bx+c$ for which $a+b=c$ will share one root if $aneq c$ (and a double root if $a=c$).

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
  – Nicholas Stull
  Nov 21 '18 at 21:14
  
  
  

  
  
  
  

add a comment | 

1

@Nicholas Stull

The quadratic equations $y=x^2+3x+2$ and $y=2x^2+3x+1$ both have the root $-1$. In fact, any quadratic equation $y=ax^2+bx+c$ for which $a+b=c$ will share one root if $aneq c$ (and a double root if $a=c$).

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
  – Nicholas Stull
  Nov 21 '18 at 21:14
  
  
  

  
  
  
  

add a comment | 

1

1

1

@Nicholas Stull

The quadratic equations $y=x^2+3x+2$ and $y=2x^2+3x+1$ both have the root $-1$. In fact, any quadratic equation $y=ax^2+bx+c$ for which $a+b=c$ will share one root if $aneq c$ (and a double root if $a=c$).

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

@Nicholas Stull

The quadratic equations $y=x^2+3x+2$ and $y=2x^2+3x+1$ both have the root $-1$. In fact, any quadratic equation $y=ax^2+bx+c$ for which $a+b=c$ will share one root if $aneq c$ (and a double root if $a=c$).

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

share|cite|improve this answer

share|cite|improve this answer

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

answered Nov 21 '18 at 21:13

MoKo19MoKo19

1914

1914

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
  – Nicholas Stull
  Nov 21 '18 at 21:14
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
  – Nicholas Stull
  Nov 21 '18 at 21:14
  
  
  

  
  
  
  

Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
– Nicholas Stull
Nov 21 '18 at 21:14

Good call...My brain isn't working apparently. Should definitely sleep more before trying to think about abstract algebra. :-(
– Nicholas Stull
Nov 21 '18 at 21:14

add a comment | 

1

No. Consider the following cases over the field $F = mathbb{C}$

$f(x) = 3x^2 + 2x + 1$ and $g(x) = x^2 + 2x + 3$ do not have the same roots.

Even simpler, $f(x) = 2x+1$ and $g(x) = x+2$ do not share a root.

---

A (slightly) more interesting question:

If ${a_0,a_1,ldots,a_n}subset F$ are all distinct elements of the field $F$, do the polynomials $f(x) = a_0 x^n + ldots + a_n$ and $g(x) = a_n x^n + ldots + a_0$ ever share a single root in the field $F$?

(Distinct is required here, because otherwise you could simply choose $a_0 = a_n$, $a_1 = a_2 = cdots = a_{n-1}$ and come up with a simple example of this. Apparently this is quite simple, for quadratics, as demonstrated above...)

share|cite|improve this answer

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

add a comment | 

1

No. Consider the following cases over the field $F = mathbb{C}$

$f(x) = 3x^2 + 2x + 1$ and $g(x) = x^2 + 2x + 3$ do not have the same roots.

Even simpler, $f(x) = 2x+1$ and $g(x) = x+2$ do not share a root.

---

A (slightly) more interesting question:

If ${a_0,a_1,ldots,a_n}subset F$ are all distinct elements of the field $F$, do the polynomials $f(x) = a_0 x^n + ldots + a_n$ and $g(x) = a_n x^n + ldots + a_0$ ever share a single root in the field $F$?

(Distinct is required here, because otherwise you could simply choose $a_0 = a_n$, $a_1 = a_2 = cdots = a_{n-1}$ and come up with a simple example of this. Apparently this is quite simple, for quadratics, as demonstrated above...)

share|cite|improve this answer

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

add a comment | 

1

1

1

No. Consider the following cases over the field $F = mathbb{C}$

$f(x) = 3x^2 + 2x + 1$ and $g(x) = x^2 + 2x + 3$ do not have the same roots.

Even simpler, $f(x) = 2x+1$ and $g(x) = x+2$ do not share a root.

---

A (slightly) more interesting question:

If ${a_0,a_1,ldots,a_n}subset F$ are all distinct elements of the field $F$, do the polynomials $f(x) = a_0 x^n + ldots + a_n$ and $g(x) = a_n x^n + ldots + a_0$ ever share a single root in the field $F$?

(Distinct is required here, because otherwise you could simply choose $a_0 = a_n$, $a_1 = a_2 = cdots = a_{n-1}$ and come up with a simple example of this. Apparently this is quite simple, for quadratics, as demonstrated above...)

share|cite|improve this answer

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

No. Consider the following cases over the field $F = mathbb{C}$

$f(x) = 3x^2 + 2x + 1$ and $g(x) = x^2 + 2x + 3$ do not have the same roots.

Even simpler, $f(x) = 2x+1$ and $g(x) = x+2$ do not share a root.

---

A (slightly) more interesting question:

If ${a_0,a_1,ldots,a_n}subset F$ are all distinct elements of the field $F$, do the polynomials $f(x) = a_0 x^n + ldots + a_n$ and $g(x) = a_n x^n + ldots + a_0$ ever share a single root in the field $F$?

(Distinct is required here, because otherwise you could simply choose $a_0 = a_n$, $a_1 = a_2 = cdots = a_{n-1}$ and come up with a simple example of this. Apparently this is quite simple, for quadratics, as demonstrated above...)

share|cite|improve this answer

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

share|cite|improve this answer

share|cite|improve this answer

edited Nov 21 '18 at 21:14

edited Nov 21 '18 at 21:14

edited Nov 21 '18 at 21:14

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

answered Nov 21 '18 at 19:37

Nicholas StullNicholas Stull

2,715921

2,715921

add a comment | 

add a comment | 

0

What is true is that $x mapsto 1/x$ is a bijection between the nonzero roots of $f$ and $g$.

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

add a comment | 

0

What is true is that $x mapsto 1/x$ is a bijection between the nonzero roots of $f$ and $g$.

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

add a comment | 

0

0

0

What is true is that $x mapsto 1/x$ is a bijection between the nonzero roots of $f$ and $g$.

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

What is true is that $x mapsto 1/x$ is a bijection between the nonzero roots of $f$ and $g$.

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

share|cite|improve this answer

share|cite|improve this answer

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

answered Nov 21 '18 at 21:27

lhflhf

163k10167387

163k10167387

add a comment | 

add a comment | 

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