If left|begin{array}{cc}x^3+2 x^2+3 x-2 & x^2 | vedclass

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(B) Let $P(x) = \left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}ight| = a x^6+b x^5+c x^4+d x^3+e

Vedclass · Accepted Answer

$25$

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(B) Let $P(x) = \left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}ight| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$.To find $a+b+c+d+e+f$,we note that $P(1) = a+b+c+d+e+f+g$.First,calculate $P(1)$ by substituting $x=1$ into the determinant:$P(1) = \left|\begin{array}{cc}1+2+3-2 & 1+2+4 \ 1-1-2-1 & 3-2+4-2\end{array}ight| = \left|\begin{array}{cc}4 & 7 \ -3 & 3\end{array}ight| = (4)(3) - (7)(-3) = 12 + 21 = 33$.Next,calculate $g$ by substituting $x=0$ into the determinant:$g = P(0) = \left|\begin{array}{cc}-2 & 4 \ -1 & -2\end{array}ight| = (-2)(-2) - (4)(-1) = 4 + 4 = 8$.Since $P(1) = a+b+c+d+e+f+g$,we have $33 = (a+b+c+d+e+f) + 8$.Therefore,$a+b+c+d+e+f = 33 - 8 = 25$.

If $\left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \\ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}\right| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$,then $a+b+c+d+e+f$ is equal to

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