Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 12

Solution

Resolve into partial fractions: $\dfrac{3x^2+4x-5}{(x-1)^3}$

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Solution

Let:

3x2+4x−5(x−1)3=Ax−1+B(x−1)2+C(x−1)3\frac{3x^2+4x-5}{(x-1)^3}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}

Multiply by (x−1)3(x-1)^3:

3x2+4x−5=A(x−1)2+B(x−1)+C3x^2+4x-5=A(x-1)^2+B(x-1)+C

Solving gives:

A=3,B=10,C=2A=3,\quad B=10,\quad C=2

Hence:

3x2+4x−5(x−1)3=3x−1+10(x−1)2+2(x−1)3\boxed{\frac{3x^2+4x-5}{(x-1)^3}=\frac{3}{x-1}+\frac{10}{(x-1)^2}+\frac{2}{(x-1)^3}}

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