Relation between roots and coefficients - manipulation of identities
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
polynomials
New contributor
New contributor
edited 2 hours ago
Eevee Trainer
6,41811237
6,41811237
New contributor
asked 2 hours ago
MmloilerMmloiler
161
161
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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}
$endgroup$
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.
$endgroup$
$begingroup$
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 |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
New contributor
$endgroup$
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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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
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}
$endgroup$
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}
$endgroup$
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}
$endgroup$
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}
answered 1 hour ago
induction601induction601
1,226314
1,226314
add a comment |
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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.
edited 1 hour ago
answered 1 hour ago
MacavityMacavity
35.5k52554
35.5k52554
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
New contributor
$endgroup$
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.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
New contributor
$endgroup$
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.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
New contributor
$endgroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
New contributor
edited 27 mins ago
Brahadeesh
6,46942363
6,46942363
New contributor
answered 1 hour ago
Wentao WangWentao Wang
11
11
New contributor
New contributor
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 |
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
5
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
$begingroup$
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 |
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
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