Further Mathematics · Lesson 5

Partial fractions & linear inequalities

WAECNECO

What the syllabus asks

WAEC: Pure Mathematics: partial fractions (distinct linear factors, repeated linear factors, quadratic factors, improper fractions); linear inequalities in one and two variables, graphical representation and linear programming; quadratic inequalities

NECO: Pure Mathematics: partial fractions (distinct linear factors, repeated linear factors, quadratic factors, improper fractions); linear inequalities in one and two variables, graphical representation and linear programming; quadratic inequalities

What you will learn

Resolve rational expressions with distinct linear factors into partial fractions.
Handle repeated linear factors and irreducible quadratic factors.
Divide improper fractions before splitting.
Solve linear inequalities in one variable and show them on a number line.
Solve quadratic and rational inequalities using critical values.
Represent inequalities in two variables graphically and solve linear programming problems.

Key points

Check degrees: if improper, divide first and keep the quotient.
Distinct linear factors: A/(x − a) + B/(x − b); use cover-up.
Repeated (x − a)²: A/(x − a) + B/(x − a)².
Irreducible quadratic: (Bx + C)/(x² + k).
Find constants by substituting roots, then compare coefficients; verify with x = 0.
Inequalities: reverse the sign when multiplying or dividing by a negative.
Quadratic inequalities: roots, then between (for < 0) or outside (for > 0) when the parabola opens up.
Linear programming: draw constraints, find vertices, test the objective function at each.

Start here: Why we split fractions

In ordinary arithmetic you add fractions: 1/2 + 1/3 = 5/6. Partial fractions is the reverse process: given a single fraction, we break it into simpler pieces. In algebra, the same idea works:

2/(x + 1) + 3/(x − 1) = [2(x − 1) + 3(x + 1)] / ((x + 1)(x − 1)) = (5x + 1)/((x + 1)(x − 1))

So (5x + 1)/((x + 1)(x − 1)) can be written back as 2/(x + 1) + 3/(x − 1). The pieces 2/(x + 1) and 3/(x − 1) are the partial fractions.

Why bother? Simple fractions are much easier to work with. In integration, ∫ 2/(x + 1) dx is easy, but ∫ (5x + 1)/(x² − 1) dx is not obvious. In series, 1/(k(k + 1)) splits into 1/k − 1/(k + 1), which makes long sums collapse. Partial fractions appear every year in WAEC Further Maths Paper 1 and frequently in Paper 2.

Worked example 5.1

Express 3/(x + 2) − 1/(x − 1) as a single fraction.

Common denominator: (x + 2)(x − 1).
Numerator: 3(x − 1) − 1(x + 2) = 3x − 3 − x − 2 = 2x − 5.

(2x − 5)/((x + 2)(x − 1))

Keep this example in mind: in 16.2 you will learn to go backwards, from (2x − 5)/((x + 2)(x − 1)) to 3/(x + 2) − 1/(x − 1).

⚠️ Common mistake: trying to split an improper fraction directly. Check degrees first (16.5).

Try it yourself. Combine 1/x − 1/(x + 1).

Show solution

[(x + 1) − x]/(x(x + 1)) = 1/(x(x + 1)).

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Practice questions

1. Resolve (5x − 1)/((x + 1)(x − 2)).

Show worked answer

x = −1: A = −6/−3 = 2; x = 2: B = 9/3 = 3. 2/(x + 1) + 3/(x − 2).

2. Resolve (x + 3)/(x + 1)².

Show worked answer

x + 3 = A(x + 1) + B; B = 2, A = 1. 1/(x + 1) + 2/(x + 1)².

3. Resolve 3/((x − 1)(x² + 2)).

Show worked answer

A = 3/3 = 1; x²: 0 = 1 + B, B = −1; constant: 3 = 2 − C, C = −1. 1/(x − 1) − (x + 1)/(x² + 2).

4. Resolve (x² + 1)/((x − 1)(x + 2)).

Show worked answer

Quotient 1, remainder −x + 3. Split: x = 1 gives 2/3; x = −2 gives 5/(−3). 1 + 2/(3(x − 1)) − 5/(3(x + 2)).

5. Solve 7 − 3x ≤ 1 + x.

Show worked answer

6 ≤ 4x → x ≥ 1.5.

Key words

Proper fraction: A rational expression whose numerator has lower degree than its denominator.
Improper fraction: Numerator degree equal to or greater than denominator degree.
Partial fractions: Simpler fractions whose sum equals a given rational expression.
Identity: An equation true for all values of x (symbol ≡).
Cover-up rule: A shortcut for finding constants over distinct linear factors.
Repeated factor: A factor appearing more than once, e.g. (x − 2)².
Irreducible quadratic: A quadratic that cannot be factorised over the real numbers, e.g. x² + 1.
Inequality: A statement using <, >, ≤ or ≥.
Critical values: Values where an expression equals zero or is undefined.
Feasible region: The set of points satisfying all constraints.
Objective function: The quantity to be maximised or minimised in linear programming.
Constraint: An inequality limiting the variables.

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