Mastering Partial Fractions for A-Level Mathematics

What the examiner is testing

Examiners assess your ability to decompose a rational function into simpler fractions, specifically identifying the correct form for linear and repeated linear factors in the denominator. Marks are awarded for correctly setting up the identity, accurately solving for the unknown constants, and presenting the final decomposed expression.

The method

1. Factorise the denominator: Ensure the denominator of the given fraction is fully factorised into its simplest linear or irreducible quadratic factors.
2. Determine the partial fraction form:
   • For each distinct linear factor \((ax+b)\) in the denominator, include a term \(\frac{A}{ax+b}\).
   • For each repeated linear factor \((cx+d)^n\), include terms \(\frac{B}{cx+d} + \frac{C}{(cx+d)^2} + \dots + \frac{D}{(cx+d)^n}\).
   • For each irreducible quadratic factor \((px^2+qx+r)\), include a term \(\frac{Ex+F}{px^2+qx+r}\).
3. Form an identity: Set the original fraction equal to the sum of its partial fraction forms. Multiply both sides by the original denominator to eliminate all denominators, creating an identity involving the numerators.
4. Solve for the unknown constants:
   • Substitution method: Substitute specific values of \(x\) (often the roots of the denominator factors) into the identity to make terms zero and directly solve for constants.
   • Coefficient comparison method: Expand the right-hand side of the identity and equate the coefficients of corresponding powers of \(x\) on both sides, forming a system of simultaneous equations.
   • Often, a combination of both methods is most efficient.
5. Write the final partial fraction decomposition: Substitute the calculated values of the constants back into the partial fraction form determined in step 2.

Worked example

Express \(\frac{5x-1}{(x-1)(x+2)}\) in partial fractions.

$$ \frac{5x-1}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2} $$

Multiply by \((x-1)(x+2)\):
$$ 5x-1 \equiv A(x+2) + B(x-1) $$

Substitute \(x=1\):
$$ 5(1)-1 = A(1+2) + B(1-1) $$
$$ 4 = 3A + 0B $$
$$ 4 = 3A $$
$$ A = \frac{4}{3} $$

Substitute \(x=-2\):
$$ 5(-2)-1 = A(-2+2) + B(-2-1) $$
$$ -10-1 = 0A - 3B $$
$$ -11 = -3B $$
$$ B = \frac{11}{3} $$

Therefore:
$$ \frac{5x-1}{(x-1)(x+2)} \equiv \frac{4}{3(x-1)} + \frac{11}{3(x+2)} $$

Sanity Check:
Let \(x=0\).
Original expression: \(\frac{5(0)-1}{(0-1)(0+2)} = \frac{-1}{-2} = \frac{1}{2}\)
Partial fractions: \(\frac{4}{3(0-1)} + \frac{11}{3(0+2)} = \frac{4}{-3} + \frac{11}{6} = -\frac{8}{6} + \frac{11}{6} = \frac{3}{6} = \frac{1}{2}\)
The values match.

Worked example: a harder one

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

A common mistake is to assume all factors are linear, or to incorrectly set up the irreducible quadratic form. If we tried \(\frac{A}{x-1} + \frac{B}{x^2+1}\), we would quickly run into issues as \(B\) would need to account for an \(x\) term in the numerator.

The correct form for the irreducible quadratic factor \((x^2+1)\) is \(\frac{Bx+C}{x^2+1}\).

$$ \frac{x^2+x+1}{(x-1)(x^2+1)} \equiv \frac{A}{x-1} + \frac{Bx+C}{x^2+1} $$

Multiply by \((x-1)(x^2+1)\):
$$ x^2+x+1 \equiv A(x^2+1) + (Bx+C)(x-1) $$

Substitute \(x=1\):
$$ (1)^2+(1)+1 = A((1)^2+1) + (B(1)+C)(1-1) $$
$$ 3 = A(2) + 0 $$
$$ 2A = 3 \implies A = \frac{3}{2} $$

Now, we can use coefficient comparison. Expand the right-hand side:
$$ x^2+x+1 \equiv Ax^2+A + Bx^2-Bx+Cx-C $$
$$ x^2+x+1 \equiv (A+B)x^2 + (-B+C)x + (A-C) $$

Equating coefficients of \(x^2\):
$$ 1 = A+B $$
Substitute \(A = \frac{3}{2}\):
$$ 1 = \frac{3}{2} + B \implies B = 1 - \frac{3}{2} = -\frac{1}{2} $$

Equating coefficients of \(x\):
$$ 1 = -B+C $$
Substitute \(B = -\frac{1}{2}\):
$$ 1 = -(-\frac{1}{2}) + C $$
$$ 1 = \frac{1}{2} + C \implies C = 1 - \frac{1}{2} = \frac{1}{2} $$

(As a check, equate constant terms: \(1 = A-C\). \(\frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1\). This matches.)

Substitute \(A, B, C\) back into the partial fraction form:
$$ \frac{x^2+x+1}{(x-1)(x^2+1)} \equiv \frac{3}{2(x-1)} + \frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1} $$
$$ \equiv \frac{3}{2(x-1)} + \frac{1-x}{2(x^2+1)} $$

Practice

1. Express \(\frac{7x-1}{x(x-1)}\) in partial fractions.
2. Express \(\frac{x+5}{(x+1)(x+2)}\) in partial fractions.
3. Express \(\frac{3x^2+2x+1}{(x+1)^2(x-1)}\) in partial fractions.
4. A function is defined by \(f(x) = \frac{2x^2+1}{(x-2)(x^2+4)}\). Find the partial fraction decomposition of \(f(x)\).

Answers:
1. \(\frac{1}{x} + \frac{6}{x-1}\)
2. \(\frac{4}{x+1} - \frac{3}{x+2}\)
3. \(\frac{1}{x-1} + \frac{2}{x+1} - \frac{2}{(x+1)^2}\)
4. Working for Q4:
$$ \frac{2x^2+1}{(x-2)(x^2+4)} \equiv \frac{A}{x-2} + \frac{Bx+C}{x^2+4} $$
$$ 2x^2+1 \equiv A(x^2+4) + (Bx+C)(x-2) $$
Substitute \(x=2\):
$$ 2(2)^2+1 = A((2)^2+4) + (B(2)+C)(2-2) $$
$$ 2(4)+1 = A(8) + 0 $$
$$ 9 = 8A \implies A = \frac{9}{8} $$
Expand the identity:
$$ 2x^2+1 \equiv Ax^2+4A + Bx^2-2Bx+Cx-2C $$
$$ 2x^2+1 \equiv (A+B)x^2 + (-2B+C)x + (4A-2C) $$
Equate coefficients of \(x^2\):
$$ 2 = A+B $$
$$ 2 = \frac{9}{8} + B \implies B = 2 - \frac{9}{8} = \frac{16-9}{8} = \frac{7}{8} $$
Equate constant terms:
$$ 1 = 4A-2C $$
$$ 1 = 4(\frac{9}{8}) - 2C $$
$$ 1 = \frac{9}{2} - 2C $$
$$ 2C = \frac{9}{2} - 1 = \frac{7}{2} $$
$$ C = \frac{7}{4} $$
(Check coefficients of \(x\): \(0 = -2B+C\). \(-2(\frac{7}{8}) + \frac{7}{4} = -\frac{7}{4} + \frac{7}{4} = 0\). This matches.)
$$ f(x) = \frac{9}{8(x-2)} + \frac{\frac{7}{8}x+\frac{7}{4}}{x^2+4} \equiv \frac{9}{8(x-2)} + \frac{7x+14}{8(x^2+4)} $$

The three mistakes that lose marks

1. Incorrectly setting up the form for repeated factors: Students often forget the intermediate terms for repeated linear factors.
   • Wrong: \(\frac{P}{(x+1)^2(x-2)} \equiv \frac{A}{(x+1)^2} + \frac{B}{x-2}\)
   • Right: \(\frac{P}{(x+1)^2(x-2)} \equiv \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2}\)
2. Incorrectly setting up the form for irreducible quadratic factors: Omitting the \(x\) term in the numerator.
   • Wrong: \(\frac{P}{(x-1)(x^2+4)} \equiv \frac{A}{x-1} + \frac{B}{x^2+4}\)
   • Right: \(\frac{P}{(x-1)(x^2+4)} \equiv \frac{A}{x-1} + \frac{Bx+C}{x^2+4}\)
3. Not handling improper fractions first: If the degree of the numerator is greater than or equal to the degree of the denominator, algebraic division must be performed before partial fractions.
   • Wrong: Trying to decompose \(\frac{x^3+1}{x^2-1}\) directly into partial fractions.
   • Right: Perform \((x^3+1) \div (x^2-1)\) to get \(x + \frac{x+1}{x^2-1}\), then decompose \(\frac{x+1}{x^2-1}\).

30-second recap

Partial fractions decompose complex rational functions into simpler ones, essential for integration and series expansions. Key steps involve correctly identifying the form for linear, repeated linear, and irreducible quadratic factors, then solving for constants using substitution and coefficient comparison. Always check for improper fractions first.