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Relation between roots and coefficients - manipulation of identities

Symmetric polynomials and the Newton identitiesRelation betwen coefficients and roots of a polynomialExpressing polynomial roots expression in terms of coefficientsPolynomials such that roots=coefficientsFind polynomial whose root is sum of roots of other polynomialsRelation between the roots and the coefficients of a polynomialFinding roots of cubic equationRoots of cubic equationWhat is the sum of all possible values of $gamma$How to prove relationship between coefficients and roots of a cubic (or) general polynomialRoots of polynomials and their formulae relating to coefficients

3

$begingroup$

The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$

I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.

polynomials

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edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

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$endgroup$

add a comment | 

3

$begingroup$

The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$

I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.

polynomials

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

New contributor

Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$

I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.

polynomials

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

New contributor

Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

New contributor

Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

New contributor

Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

Eevee Trainer

6,41811237

edited 2 hours ago

Eevee Trainer

6,41811237

asked 2 hours ago

MmloilerMmloiler

161

New contributor

Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

MmloilerMmloiler

161

3 Answers
3

active

oldest

votes

4

$begingroup$

Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}

share|cite|improve this answer

answered 1 hour ago

induction601induction601

1,226314

$endgroup$

add a comment | 

4

$begingroup$

Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$

--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

MacavityMacavity

35.5k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
  $endgroup$
  – Macavity
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$

share|cite|improve this answer

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
  $endgroup$
  – John Omielan
  1 hour ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}

share|cite|improve this answer

answered 1 hour ago

induction601induction601

1,226314

$endgroup$

add a comment | 

4

$begingroup$

Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}

share|cite|improve this answer

answered 1 hour ago

induction601induction601

1,226314

$endgroup$

add a comment | 

$begingroup$

Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}

share|cite|improve this answer

answered 1 hour ago

induction601induction601

1,226314

$endgroup$

share|cite|improve this answer

answered 1 hour ago

induction601induction601

1,226314

answered 1 hour ago

induction601induction601

1,226314

answered 1 hour ago

induction601induction601

1,226314

4

$begingroup$

Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$

--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

MacavityMacavity

35.5k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
  $endgroup$
  – Macavity
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$

--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

MacavityMacavity

35.5k52554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
  $endgroup$
  – Macavity
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$

--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

MacavityMacavity

35.5k52554

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

MacavityMacavity

35.5k52554

answered 1 hour ago

MacavityMacavity

35.5k52554

answered 1 hour ago

MacavityMacavity

35.5k52554

-1

$begingroup$

$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$

share|cite|improve this answer

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
  $endgroup$
  – John Omielan
  1 hour ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$

share|cite|improve this answer

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
  $endgroup$
  – John Omielan
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$

share|cite|improve this answer

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 28 mins ago

Brahadeesh

6,46942363

edited 28 mins ago

Brahadeesh

6,46942363

answered 1 hour ago

Wentao WangWentao Wang

11

New contributor

Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 1 hour ago

Wentao WangWentao Wang

11