If frac{2 x^3+x^2-5}{x^4-25}=frac{A x+B}{x^2- | vedclass

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(C) Given,$\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$ Since $x^4-25 = (x^2-5)(x^2+5)$,we have: $2 x^3+x^2-5 = (A x+B)(x^2+5)

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$(1, 0, 1)$

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(C) Given,$\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$ Since $x^4-25 = (x^2-5)(x^2+5)$,we have: $2 x^3+x^2-5 = (A x+B)(x^2+5) + (C x+1)(x^2-5)$ $2 x^3+x^2-5 = A x^3 + 5Ax + Bx^2 + 5B + Cx^3 - 5Cx + x^2 - 5$ $2 x^3+x^2-5 = x^3(A+C) + x^2(B+1) + x(5A-5C) + (5B-5)$ Equating the coefficients of $x^3, x^2, x$ and the constant term: $A+C = 2$ $B+1 = 1 \Rightarrow B = 0$ $5A-5C = 0 \Rightarrow A = C$ $5B-5 = -5 \Rightarrow 5(0)-5 = -5$ (which is consistent) Substituting $A=C$ into $A+C=2$,we get $2C=2$,so $C=1$ and $A=1$. Thus,$(A, B, C) = (1, 0, 1)$.

If $\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$,then $(A, B, C)$ equals to

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