Let $\begin{vmatrix}x & 2 & x \\x^2 & x & 6 \\x & x & 6 \end{vmatrix}= ax^4 +bx^3 +cx^2 +dx+e$
Then, the value of $5a+4b +3c +2d+ e$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : -11
We have,
$\begin{vmatrix}x & 2 & x \\x^2 & x & 6 \\x & x & 6 \end{vmatrix}$
$= \begin{vmatrix}x & 2 & x \\x^2 & x & 6 \\x-x^2 & 0 & 0 \end{vmatrix}$ [Applying $R_3→ R_3- R_2 $]
$=(x-x^2)(12-x^2)$
$= 12x-x^3-12x^2 +x^4 $
$∴a=1, b=-1, c=-12, d= 12 $ and $e= 0 $
$∴5a+4b + 3c + 2d + e= 5-4 - 36 + 24 +0 = -11 $