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How can you determine if a root of a polynomial is a repeated one?

1

$begingroup$

Let's say we have the polynomial $x^3 − 12x − 16$, and we know there are 2 roots, $x = -2$, and $x = 4$, and one of those roots appears twice.

How can we determine which of those two roots is the repeated one?

polynomials roots

share|cite|improve this question

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  is $f(x)=(x+2)^2(x-4)$ or $(x-4)^2(x+2)$?
  $endgroup$
  – Alexandros
  Mar 25 at 17:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's say we have the polynomial $x^3 − 12x − 16$, and we know there are 2 roots, $x = -2$, and $x = 4$, and one of those roots appears twice.

How can we determine which of those two roots is the repeated one?

polynomials roots

share|cite|improve this question

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  is $f(x)=(x+2)^2(x-4)$ or $(x-4)^2(x+2)$?
  $endgroup$
  – Alexandros
  Mar 25 at 17:16
  

  
  
  
  

add a comment | 

$begingroup$

Let's say we have the polynomial $x^3 − 12x − 16$, and we know there are 2 roots, $x = -2$, and $x = 4$, and one of those roots appears twice.

How can we determine which of those two roots is the repeated one?

polynomials roots

share|cite|improve this question

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

$endgroup$

share|cite|improve this question

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

share|cite|improve this question

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

edited Mar 25 at 17:20

Brian

1,503416

edited Mar 25 at 17:20

Brian

1,503416

asked Mar 25 at 17:13

JBrahaJBraha

1116

asked Mar 25 at 17:13

JBrahaJBraha

1116

3 Answers
3

active

oldest

votes

2

$begingroup$

You could use Vieta's formulas.

For this polynomial, the sum of the roots must be $0,$ since that is the coefficient of $x^2$;

$-2+-2+4=0$ and $4+-2+4ne0,$ so the answer is that $-2$ is repeated.

share|cite|improve this answer

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I accepted this solution as it is the quickest and easiest method for the example in my question. For more complex problems, the other users' methods (involving differentiation) could be more useful.
  $endgroup$
  – JBraha
  Mar 25 at 17:32
  

  
  
  
  

add a comment | 

1

$begingroup$

You differentiate your polynomial, which will give you $3x^2-12$. Since $-2$ is a root of it, it's $-2$ which is the double root.

share|cite|improve this answer

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

$endgroup$

add a comment | 

1

$begingroup$

With algebra, you can factor (or do something equivalent, like synthetic division or polynomial long division). Like here you can compute that $frac{x^3-12x-16}{(x+2)(x-4)}=x+2$ for $x neq -2,4$.

Another option with algebra is guess-and-check: you can multiply out $(x+2)^2(x-4)$ and $(x-4)^2(x+2)$ and see which one matches.

With calculus, you can take the derivative and plug in the roots; if you get zero at a root, then the point is at least a double root. Higher roots can be checked by taking more derivatives (e.g. $p(0)=p'(0)=p''(0)=0$ with $p'''(0) neq 0$ means a triple root).

share|cite|improve this answer

answered Mar 25 at 17:15

IanIan

69.2k25393

$endgroup$

add a comment | 

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2

$begingroup$

You could use Vieta's formulas.

For this polynomial, the sum of the roots must be $0,$ since that is the coefficient of $x^2$;

$-2+-2+4=0$ and $4+-2+4ne0,$ so the answer is that $-2$ is repeated.

share|cite|improve this answer

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I accepted this solution as it is the quickest and easiest method for the example in my question. For more complex problems, the other users' methods (involving differentiation) could be more useful.
  $endgroup$
  – JBraha
  Mar 25 at 17:32
  

  
  
  
  

add a comment | 

2

$begingroup$

You could use Vieta's formulas.

For this polynomial, the sum of the roots must be $0,$ since that is the coefficient of $x^2$;

$-2+-2+4=0$ and $4+-2+4ne0,$ so the answer is that $-2$ is repeated.

share|cite|improve this answer

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I accepted this solution as it is the quickest and easiest method for the example in my question. For more complex problems, the other users' methods (involving differentiation) could be more useful.
  $endgroup$
  – JBraha
  Mar 25 at 17:32
  

  
  
  
  

add a comment | 

$begingroup$

You could use Vieta's formulas.

For this polynomial, the sum of the roots must be $0,$ since that is the coefficient of $x^2$;

$-2+-2+4=0$ and $4+-2+4ne0,$ so the answer is that $-2$ is repeated.

share|cite|improve this answer

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

$endgroup$

share|cite|improve this answer

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

answered Mar 25 at 17:20

J. W. TannerJ. W. Tanner

5,0751520

1

$begingroup$

You differentiate your polynomial, which will give you $3x^2-12$. Since $-2$ is a root of it, it's $-2$ which is the double root.

share|cite|improve this answer

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

$endgroup$

add a comment | 

1

$begingroup$

You differentiate your polynomial, which will give you $3x^2-12$. Since $-2$ is a root of it, it's $-2$ which is the double root.

share|cite|improve this answer

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

$endgroup$

add a comment | 

$begingroup$

You differentiate your polynomial, which will give you $3x^2-12$. Since $-2$ is a root of it, it's $-2$ which is the double root.

share|cite|improve this answer

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

$endgroup$

share|cite|improve this answer

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

answered Mar 25 at 17:15

José Carlos SantosJosé Carlos Santos

176k24136245

1

$begingroup$

With algebra, you can factor (or do something equivalent, like synthetic division or polynomial long division). Like here you can compute that $frac{x^3-12x-16}{(x+2)(x-4)}=x+2$ for $x neq -2,4$.

Another option with algebra is guess-and-check: you can multiply out $(x+2)^2(x-4)$ and $(x-4)^2(x+2)$ and see which one matches.

With calculus, you can take the derivative and plug in the roots; if you get zero at a root, then the point is at least a double root. Higher roots can be checked by taking more derivatives (e.g. $p(0)=p'(0)=p''(0)=0$ with $p'''(0) neq 0$ means a triple root).

share|cite|improve this answer

answered Mar 25 at 17:15

IanIan

69.2k25393

$endgroup$

add a comment | 

1

$begingroup$

With algebra, you can factor (or do something equivalent, like synthetic division or polynomial long division). Like here you can compute that $frac{x^3-12x-16}{(x+2)(x-4)}=x+2$ for $x neq -2,4$.

Another option with algebra is guess-and-check: you can multiply out $(x+2)^2(x-4)$ and $(x-4)^2(x+2)$ and see which one matches.

With calculus, you can take the derivative and plug in the roots; if you get zero at a root, then the point is at least a double root. Higher roots can be checked by taking more derivatives (e.g. $p(0)=p'(0)=p''(0)=0$ with $p'''(0) neq 0$ means a triple root).

share|cite|improve this answer

answered Mar 25 at 17:15

IanIan

69.2k25393

$endgroup$

add a comment | 

$begingroup$

With algebra, you can factor (or do something equivalent, like synthetic division or polynomial long division). Like here you can compute that $frac{x^3-12x-16}{(x+2)(x-4)}=x+2$ for $x neq -2,4$.

Another option with algebra is guess-and-check: you can multiply out $(x+2)^2(x-4)$ and $(x-4)^2(x+2)$ and see which one matches.

With calculus, you can take the derivative and plug in the roots; if you get zero at a root, then the point is at least a double root. Higher roots can be checked by taking more derivatives (e.g. $p(0)=p'(0)=p''(0)=0$ with $p'''(0) neq 0$ means a triple root).

share|cite|improve this answer

answered Mar 25 at 17:15

IanIan

69.2k25393

$endgroup$

share|cite|improve this answer

answered Mar 25 at 17:15

IanIan

69.2k25393

answered Mar 25 at 17:15

IanIan

69.2k25393

answered Mar 25 at 17:15

IanIan

69.2k25393