Improper fractions: divide first

A fraction is improper when the degree of the numerator is at least the degree of the denominator. Divide first: \(\dfrac{\mathrm{P}(x)}{\mathrm{D}(x)}=\mathrm{Q}(x)+\dfrac{\mathrm{R}(x)}{\mathrm{D}(x)}\) with \(\deg\mathrm{R}<\deg\mathrm{D}\), then express \(\dfrac{\mathrm{R}(x)}{\mathrm{D}(x)}\) in partial fractions.

Why: every partial-fraction form has a numerator of lower degree than its denominator, so the fractions always add up to a proper fraction. The polynomial part \(\mathrm{Q}(x)\) has to be taken out first.

Method

1. Compare degrees. If the fraction is improper, multiply out the denominator.
2. Divide by long division to find the quotient \(\mathrm{Q}(x)\) and the remainder \(\mathrm{R}(x)\).
3. Express \(\dfrac{\mathrm{R}(x)}{\mathrm{D}(x)}\) in partial fractions, using the form for its denominator.
4. Write the answer as the quotient plus the partial fractions.

Example

Express \(\dfrac{x^3+2x^2-3}{x(x+1)}\) in partial fractions.

The denominator is \(x^2+x\). \(x^3\div x^2=x\): \(x(x^2+x)=x^3+x^2\), and \((x^3+2x^2+0x-3)-(x^3+x^2)=x^2+0x-3\).

\(x^2\div x^2=1\): \(1(x^2+x)=x^2+x\), and \((x^2+0x-3)-(x^2+x)=-x-3\). So the quotient is \(x+1\) and the remainder is \(-x-3\).

\(\dfrac{-x-3}{x(x+1)}=\dfrac{A}{x}+\dfrac{B}{x+1}\): \(-x-3=A(x+1)+Bx\); \(x=0\): \(A=-3\); \(x=-1\): \(-2=-B\), \(B=2\).

\(\therefore\ \dfrac{x^3+2x^2-3}{x(x+1)}=x+1-\dfrac{3}{x}+\dfrac{2}{x+1}\). Check at \(x=1\): \(\dfrac02=0\) and \(2-3+1=0\).