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3. Improper Rational Functions0%

Partial Fractions · Topic 3 of 3

3. Improper Rational Functions

Video coming soon2 worked examples

Theory

A rational function is improper when the degree of the numerator is greater than or equal to the degree of the denominator. You must divide first — using algebraic long division — to get a polynomial quotient plus a proper remainder fraction, and only then apply partial fractions to the remainder.

The Golden Rule: if the top degree is greater than or equal to the bottom degree, divide before you decompose. The final answer is a polynomial plus proper partial fractions.

⚠️ Common Examiner Traps

  • Skipping the division: setting up partial fractions directly on an improper fraction does not work — divide first.
  • Losing the quotient: the polynomial part of the answer is worth marks — don't drop it.
  • Missing-term slips: when dividing, write placeholder terms like 0x20x^2 so columns stay lined up.

Worked examples

Example 1

Express x3+2x2+3x2−1\dfrac{x^3 + 2x^2 + 3}{x^2 - 1} in partial fractions.

Step 1: The fraction is improper (degree 3 over degree 2), so divide first. Insert a placeholder for the missing xx term, then divide the leading terms: x3÷x2=xx^3 \div x^2 = x.

x(x2−1)=x3−x  ⟹  (x3+2x2+0x)−(x3−x)=2x2+xx(x^2-1) = x^3 - x \quad\implies\quad (x^3 + 2x^2 + 0x) - (x^3 - x) = 2x^2 + x

Step 2: Bring down the +3+3 to give 2x2+x+32x^2 + x + 3. Now 2x2÷x2=22x^2 \div x^2 = 2:

2(x2−1)=2x2−2  ⟹  (2x2+x+3)−(2x2−2)=x+52(x^2-1) = 2x^2 - 2 \quad\implies\quad (2x^2 + x + 3) - (2x^2 - 2) = x + 5

Step 3: The remainder x+5x+5 has lower degree than x2−1x^2-1, so the division is complete with quotient x+2x+2:

x3+2x2+3x2−1=x+2+x+5x2−1\frac{x^3+2x^2+3}{x^2-1} = x + 2 + \frac{x+5}{x^2-1}

Step 4: Decompose the proper remainder, using x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1):

x+5(x−1)(x+1)=Ax−1+Bx+1  ⟹  x+5=A(x+1)+B(x−1)\frac{x+5}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \implies x+5 = A(x+1) + B(x-1)
x=1: 6=2A  ⟹  A=3x=−1: 4=−2B  ⟹  B=−2x=1:\ 6 = 2A \implies A = 3 \qquad x=-1:\ 4 = -2B \implies B = -2

Step 5: Combine the quotient and the partial fractions:

x3+2x2+3x2−1=x+2+3x−1−2x+1\frac{x^3+2x^2+3}{x^2-1} = x + 2 + \frac{3}{x-1} - \frac{2}{x+1}

Example 2

Express 2x2+xx+1\dfrac{2x^2 + x}{x + 1} in the form of a polynomial plus a proper fraction.

Step 1: Degree 2 over degree 1 is improper, so divide. Dividing 2x2+x2x^2 + x by x+1x+1 gives quotient 2x−12x - 1 with remainder 11.

Step 2: The denominator is already a single linear factor, so no further decomposition is needed:

2x2+xx+1=2x−1+1x+1\frac{2x^2 + x}{x + 1} = 2x - 1 + \frac{1}{x+1}