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(A) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the relationship between the coefficients and the zeros $\alpha, \beta, \gamm

Vedclass · Accepted Answer

$-17$

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(A) For a cubic polynomial of the form $p(x) = ax^{3} + bx^{2} + cx + d$,the relationship between the coefficients and the zeros $\alpha, \beta, \gamma$ is given by Vieta's formulas.Specifically,the sum of the product of zeros taken two at a time is given by $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.Comparing the given polynomial $p(x) = x^{3} + x^{2} - 17x + 15$ with the standard form,we have $a = 1$,$b = 1$,$c = -17$,and $d = 15$.Substituting these values into the formula:$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-17}{1} = -17$.

If $\alpha, \beta$ and $\gamma$ are the zeros of the cubic polynomial $p(x) = x^{3} + x^{2} - 17x + 15$,then $\alpha\beta + \beta\gamma + \gamma\alpha = \dots$

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