Alg5.1.14

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Resolve \frac{x^4}{(x-1)(x-2)}\, into partial fractions.

The given expression is an improper fraction.In this case,write the remainder proper fraction as partial fractions dividing the numerator by denominator.

If we divide, we get the remainder and quotient as 15x-14,x^2+3x+7\, respectively.

Hence, \frac{x^4}{(x-1)(x-2)}=x^2+3x+7+\frac{15x-14}{x^2-3x+2}=(x^2+3x+7)+\frac{A}{x-1}+\frac{B}{x-2}\,

That is \frac{15x-14}{x^2-3x+2}=\frac{A}{x-1}+\frac{B}{x-2}\,

Equating numerators on both sides,we get 15x-14=A(x-2)+B(x-1)\,

Susbstituting x=1,x=2\,

A=-1,B=16\,

Hence \frac{x^4}{(x-1)(x-2)}=(x^2+3x+7)-\frac{1}{x-1}+\frac{16}{x-2}\,


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