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Determinant

.Let x 2 ...

Question

.Let $∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣$ = $a x^{4} + b x^{3} + c x^{2} + d x + e$ ,then the value of 5a + 4b + 3c + 2d + e is
equal to

A

0

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B

16

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C

-16

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D

-11

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Solution

The correct option is D -11
Put x=0
$∣ ∣ ∣ ∣ \begin{matrix} 0 & 2 & 0 6 & 0 & 0 0 & 0 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0 = e$
Diff. the determinant w.r.t. x
$∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x 2 x & 1 & 0 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣$
Put $x = 0$
$0 + 0 + 12 = 12 = d$
Diff. again w.r.t. x
$∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 0 x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 2 x & 1 & 0 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 x^{2} & x & 6 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 2 x & 1 & 0 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x 2 & 0 & 0 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x 2 x & 1 & 0 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 x^{2} & x & 6 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x 2 x & 1 & 0 1 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 0 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣$
Put $x = 0$
$C = - 12 S i m i l a r l y a = 1 b = - 1$
$5 - 4 - 36 + 24 = - 11$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} x & 2 & x x^{2} & x & 6 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ = A x^{4} + B x^{3} + C x^{2} + D x + E

. then the value of

5 A + 4 B + 3 C + 2 D + E

|\begin{matrix} x & 2 & x \\ x^{2} & x & 6 \\ x & x & 6 \end{matrix}| = a x^{4} + b x^{3} + c x^{2} + d x + e

Then, the value of

5 a + 4 b + 3 c + 2 d + e

is equal to
(a) 0
(b) − 16
(c) 16
(d) none of these

2, 5, 7, - 4

are the roots of

a x^{4} + b x^{3} + c x^{2} + d x + e = 0

then the roots of

a x^{4} - b x^{3} + c x^{2} - d x + e = 0

x + 1

a x^{4} + b x^{3} + c x^{2} + d x + e = 0

What is the degree of the polynomial p (x) : p (x) = a x^{4} + b x^{3} + c x^{2} + d x + e, given \frac{a}{b} = 0 & b \neq 0.

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