The partial fraction of frac{x^2}{x^2+3x-4} i | vedclass

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Question

(A) Given the expression $\frac{x^2}{x^2+3x-4}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$

Vedclass · Accepted Answer

$1+\frac{-16}{5(x+4)}+\frac{1}{5(x-1)}$

Vedclass · Answer

(A) Given the expression $\frac{x^2}{x^2+3x-4}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$\frac{x^2}{x^2+3x-4} = 1 - \frac{3x-4}{x^2+3x-4} = 1 + \frac{-3x+4}{(x-1)(x+4)}$.Let $\frac{-3x+4}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}$.Then $-3x+4 = A(x+4) + B(x-1)$.Comparing coefficients:$A+B = -3$ and $4A-B = 4$.Adding the two equations: $5A = 1 \Rightarrow A = \frac{1}{5}$.Substituting $A$ into $A+B = -3$: $\frac{1}{5} + B = -3 \Rightarrow B = -3 - \frac{1}{5} = -\frac{16}{5}$.Thus,$\frac{x^2}{x^2+3x-4} = 1 + \frac{1}{5(x-1)} - \frac{16}{5(x+4)}$.

The partial fraction of $\frac{x^2}{x^2+3x-4}$ is

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