Question

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Answer 1

Question :-

$\sf If \: \dfrac{3x^3-2x-1}{x^4+x^2+1} = \dfrac{Ax + B}{x^2 + x + 1} + \dfrac{Cx + D}{x^2 + kx +1} \: then\: k =$

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Answer :-

Firstly, Solving the RHS part :-

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➩ $\sf \dfrac{Ax + B}{x^2 + x + 1} + \dfrac{Cx + D}{x^2 + kx +1}$

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➩ $\sf \dfrac{[Ax + B](x^2 + kx + 1) + [Cx + D](x^2+x+1)}{[x^2 + x +1](x^2 + kx + 1)}$

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➩ $\sf \dfrac{Ax^3+Akx^2+Ax + Bx^2 + Bkx + B + Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D}{x^4 + kx^3 + x^2 + x^3 + kx^2 + x + x^2 + kx + 1}$

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➩ $\sf \dfrac{(Ax^3 + Cx^3)+(Akx^2 + Bx^2 + Cx^2 + Dx^2)+(Ax + Bkx + Cx + Dx)+(B+D)}{x^4 + (kx^3 + x^3) + (kx^2 + 2x^2) + (kx + x) + 1}$

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➩ $\sf \dfrac{(A+C)x^3 + (Ax + Bx + C + D)x^2 + (A + Bk + C + D)x + ( B + D )}{x^4 + ( k + 1 )x^3 + ( k + 2)x^2 + (k+1)x + 1}$

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➩ $\sf LHS = \dfrac{3x^3-2x-1}{x^4+x^2+1}$

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Comparing LHS and RHS

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➩ $\sf \dfrac{3x^3-2x-1}{x^4+x^2+1} = \dfrac{(A+C)x^3 + (Ax + Bx + C + D)x^2 + (A + Bk + C + D)x + ( B + D )}{x^4 + ( k + 1 )x^3 + ( k + 2)x^2 + (k+1)x + 1}$

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Comparing the denominator :-

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➩ $\sf x^4+x^2+1 = x^4 + ( k + 1 )x^3 + ( k + 2)x^2 + (k+1)x + 1$

➩ $\sf k + 1 = 0$

➩ $\boxed{\boxed{\sf k = -1 }}$