If x^{3}-2x^{2}-9x+18=0 and A=left|begin{arra | vedclass

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Question

(A) Given the equation $x^{3}-2x^{2}-9x+18=0$.Factoring the equation: $x^{2}(x-2)-9(x-2)=0 \Rightarrow (x^{2}-9)(x-2)=0 \Rightarrow (x-3)(x+3)(x-2)=0$

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(A) Given the equation $x^{3}-2x^{2}-9x+18=0$.Factoring the equation: $x^{2}(x-2)-9(x-2)=0 \Rightarrow (x^{2}-9)(x-2)=0 \Rightarrow (x-3)(x+3)(x-2)=0$.Thus,the possible values for $x$ are $x=2, 3, -3$.Now,evaluate the determinant $A = \left|\begin{array}{lll}1 & 2 & 3 \ 4 & x & 6 \ 7 & 8 & 9\end{array}ight|$.Expanding along the first row: $A = 1(9x - 48) - 2(36 - 42) + 3(32 - 7x)$.$A = 9x - 48 - 2(-6) + 96 - 21x$.$A = 9x - 48 + 12 + 96 - 21x = -12x + 60$.Now,substitute the values of $x$:For $x=2$: $A = -12(2) + 60 = -24 + 60 = 36$.For $x=3$: $A = -12(3) + 60 = -36 + 60 = 24$.For $x=-3$: $A = -12(-3) + 60 = 36 + 60 = 96$.Comparing the values $36, 24, 96$,the maximum value of $A$ is $96$.

If $x^{3}-2x^{2}-9x+18=0$ and $A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$,then the maximum value of $A$ is

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