What is constant numerator over a quadratic?

An irreducible quadratic factor needs a linear numerator Bx + C, not just a constant.

State the correct partial-fraction form for 1(x + 3)(x^2 + 1). [1 mark]

State the form for x(x - 2)^2(x + 1). [2 marks]

Express 4(x - 1)^2 in partial fractions. [2 marks]

SEAB Original-style practice: Express 3x + 5(x - 1)(x + 2)^2 in partial fractions.

The repeated factor (x + 2)^2 needs two fractions: Ax - 1 + Bx + 2 + C(x + 2)^2. Clear denominators: 3x + 5 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1). Let x = 1: 8 = A(9), so A = 89. Let x = -2: -1 = C(-3), so C = 13. Compare x^2 terms: 0 = A + B, so B = -89. So the result is 8/9x - 1 - 8/9x + 2 + 1/3(x + 2)^2. Markers reward the correct repeated-factor form, the two substitutions, comparing coefficients for B, and all three constants.

SEAB Original-style practice: Express 2x^2 + x + 3(x + 1)(x^2 + 2) in partial fractions.

The irreducible quadratic needs a linear numerator: Ax + 1 + Bx + Cx^2 + 2. Clear denominators: 2x^2 + x + 3 = A(x^2 + 2) + (Bx + C)(x + 1). Let x = -1: 2 - 1 + 3 = A(3), so A = 43. Compare x^2: 2 = A + B, so B = 23. Compare constants: 3 = 2A + C, so C = 3 - 83 = 13. So 4/3x + 1 + (2/3)x + 1/3x^2 + 2. Markers reward the linear numerator over the quadratic, one substitution, comparing coefficients, and all constants.

SingaporeAdditional MathematicsSyllabus dot point

How does the partial-fraction form change when the denominator has a repeated factor or an irreducible quadratic factor?

Express proper fractions with repeated linear factors or an irreducible quadratic factor as partial fractions, choosing the correct numerator forms

A focused answer to the O-Level A-Maths outcome on harder partial fractions. The correct numerator forms for a repeated linear factor and an irreducible quadratic factor, and finding the constants.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to handle the two harder denominators in partial fractions: a repeated linear factor and an irreducible quadratic factor. Each demands a particular numerator form, and getting that form right is half the battle. The constants are then found by a mix of root substitution and comparing coefficients.

The answer

A repeated linear factor

A factor raised to a power, such as $(x + 2)^2$ , contributes one fraction for each power up to that power:

\frac{px + q}{(x - a)(x + 2)^2} = \frac{A}{x - a} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}.

You need a separate term for $(x + 2)$ and for $(x + 2)^2$ ; a single fraction is not enough.

An irreducible quadratic factor

A quadratic factor that does not factorise (its discriminant is negative), such as $x^2 + 2$ , needs a linear numerator:

\frac{\dots}{(x - a)(x^2 + 2)} = \frac{A}{x - a} + \frac{Bx + C}{x^2 + 2}.

The numerator over a quadratic is one degree lower, hence $Bx + C$ , not a constant.

Finding the constants

Clear all denominators to get an identity. Substituting the real roots gives some constants directly. The remaining constants come from comparing coefficients of matching powers of $x$ on both sides (the coefficient of $x^2$ , the constant term, and so on).

Why the right form matters

If you use a constant numerator over a quadratic, or only one fraction for a repeated factor, the identity has no solution. The numerator forms are not optional; they are forced by the algebra.

Examples in context

Example 1. Integrating a repeated-root fraction. A fraction with $(x - a)^2$ in the denominator integrates to a logarithm plus a reciprocal term; the two partial fractions correspond exactly to those two integral pieces, which is why the split is needed before integrating.

Example 2. Resonance terms in models. An irreducible quadratic factor in a model often signals oscillatory behaviour, and its linear-over-quadratic partial fraction separates the steady part from the oscillating part, a meaningful split in applied work.

Try this

Q1. State the correct partial-fraction form for $\dfrac{1}{(x + 3)(x^2 + 1)}$ . [1 mark]

Cue. $\dfrac{A}{x + 3} + \dfrac{Bx + C}{x^2 + 1}$ .

Q2. State the form for $\dfrac{x}{(x - 2)^2(x + 1)}$ . [2 marks]

Cue. $\dfrac{A}{x - 2} + \dfrac{B}{(x - 2)^2} + \dfrac{C}{x + 1}$ .

Q3. Express $\dfrac{4}{(x - 1)^2}$ in partial fractions. [2 marks]

Cue. Already a single repeated term: $\dfrac{0}{x - 1} + \dfrac{4}{(x - 1)^2}$ , that is just $\dfrac{4}{(x - 1)^2}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksExpress

\dfrac{3x + 5}{(x - 1)(x + 2)^2}

in partial fractions.

Show worked answer →

The repeated factor $(x + 2)^2$ needs two fractions: $\dfrac{A}{x - 1} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^2}$ .

Clear denominators: $3x + 5 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1)$ .

Let $x = 1$ : $8 = A(9)$ , so $A = \dfrac{8}{9}$ . Let $x = -2$ : $-1 = C(-3)$ , so $C = \dfrac{1}{3}$ .

Compare $x^2$ terms: $0 = A + B$ , so $B = -\dfrac{8}{9}$ .

So the result is $\dfrac{8/9}{x - 1} - \dfrac{8/9}{x + 2} + \dfrac{1/3}{(x + 2)^2}$ .

Markers reward the correct repeated-factor form, the two substitutions, comparing coefficients for $B$ , and all three constants.

Original6 marksExpress

\dfrac{2x^2 + x + 3}{(x + 1)(x^2 + 2)}

in partial fractions.

Show worked answer →

The irreducible quadratic needs a linear numerator: $\dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 2}$ .

Clear denominators: $2x^2 + x + 3 = A(x^2 + 2) + (Bx + C)(x + 1)$ .

Let $x = -1$ : $2 - 1 + 3 = A(3)$ , so $A = \dfrac{4}{3}$ .

Compare $x^2$ : $2 = A + B$ , so $B = \dfrac{2}{3}$ . Compare constants: $3 = 2A + C$ , so $C = 3 - \dfrac{8}{3} = \dfrac{1}{3}$ .

So $\dfrac{4/3}{x + 1} + \dfrac{(2/3)x + 1/3}{x^2 + 2}$ .

Markers reward the linear numerator over the quadratic, one substitution, comparing coefficients, and all constants.