If alpha is a real root of the equation x^3+6 | vedclass

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(C) First,find the real root of the equation $x^3+6x^2+5x-42=0$. By testing integer factors of $-42$,we find that for $x=2$: $(2)^3+6(2)^2+5(2)-42 = 8

Vedclass · Accepted Answer

$-105$

Vedclass · Answer

(C) First,find the real root of the equation $x^3+6x^2+5x-42=0$. By testing integer factors of $-42$,we find that for $x=2$: $(2)^3+6(2)^2+5(2)-42 = 8+24+10-42 = 0$. Thus,$\alpha=2$ is a root.Substitute $\alpha=2$ into the matrix:$M = \left[\begin{array}{ccc}2-1 & 2+1 & 2+2 \ 2-2 & 2+3 & 2-3 \ 2+4 & 2-4 & 2+5\end{array}ight] = \left[\begin{array}{ccc}1 & 3 & 4 \ 0 & 5 & -1 \ 6 & -2 & 7\end{array}ight]$.Now,calculate the determinant $|M|$:$|M| = 1(5 	imes 7 - (-1) 	imes (-2)) - 3(0 	imes 7 - (-1) 	imes 6) + 4(0 	imes (-2) - 5 	imes 6)$$|M| = 1(35 - 2) - 3(0 + 6) + 4(0 - 30)$$|M| = 1(33) - 3(6) + 4(-30)$$|M| = 33 - 18 - 120 = -105$.

If $\alpha$ is a real root of the equation $x^3+6x^2+5x-42=0$,then the determinant of the matrix $\left[\begin{array}{ccc}\alpha-1 & \alpha+1 & \alpha+2 \\ \alpha-2 & \alpha+3 & \alpha-3 \\ \alpha+4 & \alpha-4 & \alpha+5\end{array}\right]$ is

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