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(D) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2-10x-24=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$,$\alpha\beta+\beta\gamma+\gamma\

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$28$

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(D) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2-10x-24=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$,$\alpha\beta+\beta\gamma+\gamma\alpha = -10$,and $\alpha\beta\gamma = 24$. Let $S = \alpha+\beta+\gamma = -3$. Then the roots of the new equation are $\alpha(S-\alpha) = \alpha S - \alpha^2$,$\beta S - \beta^2$,and $\gamma S - \gamma^2$. We need to find $q$,which is the sum of the products of these roots taken two at a time: $q = (\alpha S - \alpha^2)(\beta S - \beta^2) + (\beta S - \beta^2)(\gamma S - \gamma^2) + (\gamma S - \gamma^2)(\alpha S - \alpha^2)$. Expanding this: $q = S^2(\alpha\beta+\beta\gamma+\gamma\alpha) - S(\alpha\beta(\alpha+\beta) + \beta\gamma(\beta+\gamma) + \gamma\alpha(\gamma+\alpha)) + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2)$. Using $\alpha\beta(\alpha+\beta) = \alpha\beta(S-\gamma) = S\alpha\beta - \alpha\beta\gamma$,we get: $q = S^2(\sum \alpha\beta) - S(S \sum \alpha\beta - 3\alpha\beta\gamma) + ((\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha))$. Substituting values: $q = (-3)^2(-10) - (-3)((-3)(-10) - 3(24)) + ((-10)^2 - 2(24)(-3))$. $q = -90 + 3(30 - 72) + (100 + 144)$. $q = -90 + 3(-42) + 244 = -90 - 126 + 244 = 28$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-10x-24=0$,and $\alpha(\beta+\gamma), \beta(\gamma+\alpha), \gamma(\alpha+\beta)$ are the roots of the equation $x^3+px^2+qx+r=0$,then $q=$

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