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

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

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$(1, 0, 1)$

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(C) Given, $\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$ Since $x^4-25 = (x^2-5)(x^2+5)$, we have: $2x^3+x^2-5 = (Ax+B)(x^2+5) + (Cx+1)(x^2-5)$ $2x^3+x^2-5 = Ax^3 + 5Ax + Bx^2 + 5B + Cx^3 - 5Cx + x^2 - 5$ $2x^3+x^2-5 = x^3(A+C) + x^2(B+1) + x(5A-5C) + (5B-5)$ Equating the coefficients of $x^3, x^2, x$ and the constant term: $1) A+C = 2$ $2) B+1 = 1 \Rightarrow B = 0$ $3) 5A-5C = 0 \Rightarrow A = C$ $4) 5B-5 = -5 \Rightarrow 5(0)-5 = -5$ (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 $\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$, then $(A, B, C)$ equals to

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