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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.

PrincessEev
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Mmloiler
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3 Answers3

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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 $$ x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta \gamma + \gamma \alpha)x - \alpha\beta \gamma = x^3+3x^2-2x + 1 $$ that $$ \alpha+\beta+\gamma = -3, \quad \alpha\beta + \beta \gamma + \gamma \alpha = -2, \quad \alpha\beta \gamma = -1. $$ Note that $$ -2\alpha = \alpha(\alpha\beta + \beta \gamma + \gamma \alpha) = \alpha^2(\beta+\gamma) +\alpha\beta\gamma = \alpha^2(\beta+\gamma) - 1. $$ Thus $\alpha^2(\beta + \gamma)=1-2\alpha$. Similarly, $\beta^2(\alpha + \gamma) = 1-2\beta$ and $\gamma^2(\alpha + \beta) = 1-2\gamma$. Therefore, \begin{align} \alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta) &= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\ &= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9. \end{align}

induction601
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Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint: $$\sum \alpha^2(\beta+\gamma) = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma$$

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

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

Wentao Wang
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    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. – John Omielan Feb 21 '19 at 05:54
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    Some explanation of some of the workings here might help too - eg whzat identity you are using at each step. I'm having trouble seeing how you get from line 2 to 3 in your solution. I can see $(a+b+c)$ being replaced by $-3$ but I don't understand why $a^3+b^3+c^3$ becomes $(a+b+c)^3$... For a complete answer you should also probably explain things like why $a+b+c = -3$ – Chris Feb 21 '19 at 13:11