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(C) Given the cubic equation $x^3 + 3 x^2 - 10 x - 24 = 0$. By testing integer roots,for $x = 3$: $(3)^3 + 3(3)^2 - 10(3) - 24 = 27 + 27 - 30 - 24 = 0

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the cubic equation $x^3 + 3 x^2 - 10 x - 24 = 0$. By testing integer roots,for $x = 3$: $(3)^3 + 3(3)^2 - 10(3) - 24 = 27 + 27 - 30 - 24 = 0$. So,$(x - 3)$ is a factor. Dividing the polynomial by $(x - 3)$,we get $(x - 3)(x^2 + 6 x + 8) = 0$. Factoring the quadratic part: $(x - 3)(x + 4)(x + 2) = 0$. The roots are $3, -2, -4$. Given $\alpha > \beta > \gamma$,we have $\alpha = 3, \beta = -2, \gamma = -4$. Substitute these into the expression $\alpha^3 + 3 \beta^2 - 10 \gamma - 24$: $(3)^3 + 3(-2)^2 - 10(-4) - 24 = 27 + 12 + 40 - 24 = 55$. Given $55 = 11 k$,we find $k = 5$.

$\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 3 x^2 - 10 x - 24 = 0$. If $\alpha > \beta > \gamma$ and $\alpha^3 + 3 \beta^2 - 10 \gamma - 24 = 11 k$,then $k = $

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