The partial fractions of frac{x^2 - 5}{x^2 - | vedclass

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(D) Given expression: $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform lon

Vedclass · Accepted Answer

$1 + \frac{4}{x - 1} - \frac{1}{x - 2}$

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(D) Given expression: $f(x) = \frac{x^2 - 5}{x^2 - 3x + 2}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{3x - 7}{x^2 - 3x + 2}$.Now,factor the denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$.We express the remainder as partial fractions: $\frac{3x - 7}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.$3x - 7 = A(x - 2) + B(x - 1)$.Setting $x = 1$: $3(1) - 7 = A(1 - 2)$ $\Rightarrow -4 = -A$ $\Rightarrow A = 4$.Setting $x = 2$: $3(2) - 7 = B(2 - 1) \Rightarrow -1 = B$.Thus,$\frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{4}{x - 1} - \frac{1}{x - 2}$.

The partial fractions of $\frac{x^2 - 5}{x^2 - 3x + 2}$ are

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