Let - - 2 6 x- x x2 2- x2 x 3 - - - x5 - Ax4 - Bx3 - Cx2 - Dx - E- then the value of E is equal to- - - 2 6 x- x x2 2- x2 x 3 - - - x5 - Ax4 - Bx3 - Cx2 - Dx - E - - E - - - -

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Algebra of Derivatives

Let | [ 2 6...

Question

Let $∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 6 & x x & x^{2} & 2 x^{2} & x & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - x^{5} + A x^{4} + B x^{3} + C x^{2} + D x + E$ , then the value of E is equal to

माना $∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 6 & x x & x^{2} & 2 x^{2} & x & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - x^{5} + A x^{4} + B x^{3} + C x^{2} + D x + E$ , तब E का मान बराबर है

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Similar questions

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

x^{2} - 1

|\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

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

f : R \to R

f (x) = x^{2} - 3 x + 2

f (x^{2} - 3 x - 2) = a x^{4} + b x^{3} + c x^{2} + d x + e

a + b + c + d + e =

(x^{2} - 1)

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

a + c + e = b + d = 0

If x²-1 Is a factor of ax⁴+bx³+cx²+dx+e,then show that ,a+c+e=b+d=0 ?

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