A1Arena Open the app

Algebraic Expressions and Factorisation

Mathematics · WAEC and JAMB · SS2 and SS3

Factorisation and change of subject sit underneath quadratics, simultaneous equations and mensuration, so weakness here costs marks all over the paper. WASSCE tests it directly in the objectives and indirectly in nearly every theory question.

What you need to know

  • Always look for a common factor first. 12x^3y - 18x^2y^2 has 6x^2y common, giving 6x^2y(2x - 3y), and missing it makes the rest of the question much harder.
  • Factorising by grouping works when there are four terms: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y). Arrange the terms so the brackets match before you factor out.
  • The difference of two squares, a^2 - b^2 = (a - b)(a + b), is the single most examined identity. It also speeds up arithmetic: 99^2 - 1 = (99 - 1)(99 + 1) = 98 x 100 = 9,800.
  • A perfect square trinomial factors as a^2 + 2ab + b^2 = (a + b)^2, so x^2 + 10x + 25 = (x + 5)^2. Recognise it by checking that the last term is the square of half the middle coefficient.
  • To factorise ax^2 + bx + c when a is not 1, find two numbers whose PRODUCT is a x c and whose SUM is b, split the middle term with them, then group.
  • When simplifying algebraic fractions, factorise the top and bottom fully before cancelling, and only cancel whole factors. In (x^2 - 9)/(x + 3) you factorise to (x - 3)(x + 3)/(x + 3) = x - 3; you may never cancel the loose 9 against the 3.
  • To add algebraic fractions, use the lowest common denominator just as with numbers: 1/(x - 2) - 1/(x + 2) = [(x + 2) - (x - 2)]/[(x - 2)(x + 2)] = 4/(x^2 - 4).
  • Changing the subject of a formula uses the same moves as solving an equation: do the same operation to both sides, undoing in reverse order of the original operations.
  • If the required subject appears inside a root, square both sides; if it is in a denominator, multiply through by that denominator first. From T = 2 pi sqrt(l/g), squaring gives T^2 = 4 pi^2 l/g, so l = T^2 g / (4 pi^2).
  • If the new subject appears TWICE, gather every term containing it on one side, factor it out, then divide by the bracket. This is the step most candidates have never been taught.
  • Substitution questions demand brackets around negative values. If x = -3, then 2x^2 = 2(-3)^2 = 18, not -18, because the square is taken before the multiplication.
  • Expanding (a + b)^2 gives a^2 + 2ab + b^2, never a^2 + b^2. The cross term 2ab is the mark the examiner is checking for.
  • Remainder and factor theorem questions appear in further work but the idea helps here: if x = 2 makes an expression zero, then (x - 2) is one of its factors, which gives a quick way to check your factorisation.
  • Always expand your final factorisation mentally to confirm you get back the original expression; it takes five seconds and catches sign errors.

Key terms

Expression
A collection of terms joined by plus or minus signs with no equality sign.
Factorisation
The process of writing an expression as a product of two or more simpler expressions.
Like terms
Terms containing exactly the same letters raised to exactly the same powers, which alone may be added or subtracted.
Coefficient
The numerical part multiplying the variable part of a term.
Subject of a formula
The single variable standing alone on one side of a formula, expressed in terms of all the others.
Identity
An equation that is true for every value of the variable, such as (a + b)^2 = a^2 + 2ab + b^2.

Formulae

  • a^2 - b^2 = (a - b)(a + b)
  • (a + b)^2 = a^2 + 2ab + b^2
  • (a - b)^2 = a^2 - 2ab + b^2
  • a^3 + b^3 = (a + b)(a^2 - ab + b^2)
  • a^3 - b^3 = (a - b)(a^2 + ab + b^2)
  • ax + ay + bx + by = (a + b)(x + y)
  • to split ax^2 + bx + c, find two numbers with product a*c and sum b

Worked examples

Factorise completely 6x^2 - 11x - 10.

  1. Here a = 6, b = -11 and c = -10, so the product a x c = 6 x -10 = -60.
  2. Find two numbers with product -60 and sum -11. They are -15 and +4, since -15 x 4 = -60 and -15 + 4 = -11.
  3. Split the middle term: 6x^2 - 15x + 4x - 10.
  4. Group in pairs: (6x^2 - 15x) + (4x - 10).
  5. Factor each pair: 3x(2x - 5) + 2(2x - 5).
  6. The bracket (2x - 5) is common, so the answer is (3x + 2)(2x - 5).
  7. Check by expanding: 6x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10.

Make x the subject of the formula y = (2x + 3) / (x - 1).

  1. Clear the denominator by multiplying both sides by (x - 1): y(x - 1) = 2x + 3.
  2. Expand the left side: xy - y = 2x + 3.
  3. The required subject x now appears on both sides, so collect the x terms on one side: xy - 2x = 3 + y.
  4. Factor out x: x(y - 2) = y + 3.
  5. Divide both sides by (y - 2): x = (y + 3) / (y - 2).
  6. Check with a value: if x = 3 then y = (6 + 3)/2 = 4.5, and (4.5 + 3)/(4.5 - 2) = 7.5/2.5 = 3.

The mistake to avoid

In change-of-subject questions where the new subject appears twice, candidates divide too early and leave the subject on both sides, which earns nothing. The other standing error is cancelling single terms out of a fraction, writing (x + 4)/4 as x, instead of cancelling only complete factors.

In the exam

Factorise completely, not partially: if the examiner wants 2(x - 3)(x + 3) and you stop at 2(x^2 - 9), you lose the final mark. In change-of-subject questions, show the collecting and factoring lines separately because each carries its own mark, and test your result with one simple pair of numbers before you move on.