Question
) Express as a sum of partial fractions.
Answer :
Word Count : 296
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We are asked to express [ \frac{x-1}{x^3 - x^2 - 2x} ] as a sum of partial fractions. First, factor the denominator: [ x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x-2)(x+1) ] So we have: [ \frac{x-1}{x(x-2)(x+1)} ] Let us express it as: [ \frac{x-1}{x(x-2)(x+1)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} ] Multiply through by (x(x-2)(x+1)) to get: [ x-1 = A(x-2)(x+1) + B(x)(x+1) + C(x)(x-2) ] Now expand each term: 1. (A(x-2)(x+1) = A(x^2 - x - 2)) 2. (B x(x+1) = B(x^2 + x)) 3. (C x(x-2) = C(x^2 - 2x)) Combine terms: [ x-1 = (A + B + C)x^2 + (-A + B - 2C)x +
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