Algebraic Expressions: Expansion, Factorisation, Simultaneous Equations and Partial Fractions

Why algebra matters

Algebra appears all over WAEC Paper 1 and Paper 2: simplifying, expanding, factorising and solving equations. These skills build on each other, and you also need them for quadratics, indices and calculus.

Key words

In 5x² + 3x − 7:

• 5x², 3x and −7 are terms;
• 5 and 3 are coefficients, x is the variable and −7 is the constant.

Like terms have the same letters raised to the same powers (3x and 7x; 4y² and 9y²). Unlike terms do not (3x and 3y; x² and x).

Simplifying: collecting like terms

Only like terms can be added or subtracted. Add their coefficients and keep the letter part the same.

Substitution

Replace each letter with its value, then work it out using BODMAS (powers before multiplication).

Expanding brackets

Single bracket: multiply everything inside by the term outside: a(b + c) = ab + ac.

Double brackets: multiply every term in the first bracket by every term in the second (FOIL: First, Outer, Inner, Last), then collect like terms.

Special products

Learn these three patterns. They appear constantly in WAEC questions.

• (a + b)² = a² + 2ab + b², e.g. (x + 5)² = x² + 10x + 25
• (a − b)² = a² − 2ab + b², e.g. (x − 3)² = x² − 6x + 9
• Difference of two squares: (a + b)(a − b) = a² − b², e.g. (2x + 3)(2x − 3) = 4x² − 9

Factorisation: expansion in reverse

Factorising means writing an expression as a product of brackets.

1. Common factor: 6x² + 9x = 3x(2x + 3)
2. Grouping: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)
3. Difference of two squares: x² − 49 = (x − 7)(x + 7); 4x² − 25 = (2x − 5)(2x + 5)
4. Trinomials (x² + bx + c): find two numbers that multiply to c and add to b. For x² − 5x + 6, the numbers −2 and −3 multiply to 6 and add to −5, so it is (x − 2)(x − 3).

Factorising is the key skill for solving quadratic equations (next class).

Simultaneous linear equations

Two equations with two unknowns are solved together. There are two methods.

Elimination: multiply one or both equations so that one letter has the same (or opposite) coefficient, then add or subtract to remove it.

Substitution: make one letter the subject of one equation and substitute it into the other.

Word problems with simultaneous equations

Let letters stand for the unknowns, write two equations from the facts, then solve.

Partial fractions

Partial fractions reverse the process of adding fractions: one complicated fraction is written as a sum of simpler fractions. This is tested in WAEC Further Mathematics and is needed for integration.

Step 0 (proper fraction?): the degree (highest power) of the numerator must be less than that of the denominator. If not, divide first.

The three standard forms:

• Distinct linear factors: (px + q)/((x − a)(x − b)) = A/(x − a) + B/(x − b)
• Repeated linear factor: (px + q)/(x − a)² = A/(x − a) + B/(x − a)²
• Quadratic factor that won't factorise: (…)/((x − a)(x² + k)) = A/(x − a) + (Bx + C)/(x² + k)

Method: multiply both sides by the denominator, then substitute convenient values of x (the ones that make a bracket zero) and/or compare coefficients.

Practice

1. Simplify 7x + 4y − 3x + y.
2. Expand (2x + 1)(x + 3).
3. Expand (3x + 2)(3x − 2).
4. Factorise x² − 9x + 20.
5. Solve 2x + y = 11 and x − y = 1.
6. Two numbers have a sum of 27 and a difference of 5. Find them.
7. Express (x + 7)/((x − 1)(x + 2)) in partial fractions.

Answers: 1. 4x + 5y · 2. 2x² + 7x + 3 · 3. 9x² − 4 · 4. (x − 4)(x − 5) · 5. x = 4, y = 3 · 6. x + y = 27, x − y = 5, so 16 and 11 · 7. x + 7 = A(x + 2) + B(x − 1); x = 1 gives A = 8/3, x = −2 gives B = −5/3, so 8/(3(x − 1)) − 5/(3(x + 2))

Previous topic: Indices, Surds and Logarithms · Next class: Quadratic Equations.