Question

(1) X-1 (x-2)(x-3) - Partial Fractions
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Answer 1

Answer:

The expansion for x2(x−1)3 is given by

x2(x−1)3=A(x−1)+B(x−1)2+C(x−1)3

reducing the right member to common denominator we have

x2(x−1)3=A(x−1)2+B(x−1)+C(x−1)3

Equating the numerators we have

x2=Ax2+(B−2A)x+A−B+C

giving the following conditions

⎧⎪⎨⎪⎩A=1B−2A=0A−B+C=0

solving for A,B,C we get

x2(x−1)3=1(x−1)+2(x−1