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O-Level Mathematics · Secondary 3 and 4

Algebraic Expressions and Formulae

This topic covers the algebra you use in almost every O-Level Mathematics question: expanding brackets, factorising, simplifying and combining algebraic fractions, and changing the subject of a formula. Much of it begins in lower secondary and is extended in upper secondary to harder factorisation and to fractions with algebraic denominators. Fluent manipulation lets you solve quadratic equations, rearrange formulae and simplify answers in vectors and graphs without losing marks to small slips.

What the syllabus expects

  • Expand products of algebraic expressions, including (a + b)², (a − b)² and (a + b)(a − b)
  • Factorise expressions by taking out common factors, by grouping, and as quadratic expressions ax² + bx + c
  • Factorise using a² − b², a² + 2ab + b² and a² − 2ab + b²
  • Simplify, add, subtract, multiply and divide algebraic fractions, including those with algebraic denominators
  • Change the subject of a formula, including formulae where the new subject appears more than once or under a square root
  • Substitute values into a formula and evaluate it

Expanding and the special products

To expand, multiply every term in one bracket by every term in the other. (x + 4)(2x − 3) = 2x² − 3x + 8x − 12 = 2x² + 5x − 12. Three products are used so often that it pays to know them: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and (a + b)(a − b) = a² − b². So (2x − 3)² = 4x² − 12x + 9, not 4x² + 9. The middle term 2ab is the one students most often forget. When a bracket is multiplied by a negative number, every sign inside changes: −2(x − 5) = −2x + 10. Be especially careful when an expression such as 3 − (x − 2)(x + 1) is expanded. Expand the product first, keeping it in a bracket, then remove the bracket: 3 − (x² − x − 2) = 3 − x² + x + 2 = 5 + x − x².

Factorising

Factorising reverses expanding. Always take out the highest common factor first: 10x² − 4x = 2x(5x − 2). With four terms, try grouping in pairs: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y). For a quadratic expression ax² + bx + c, look for two numbers that multiply to give a × c and add to give b, then split the middle term. For 2x² + 7x + 3, a × c = 6, and 6 and 1 add to 7, so 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Spot the difference of two squares quickly: 9x² − 16 = (3x − 4)(3x + 4), and 2x² − 50 = 2(x² − 25) = 2(x − 5)(x + 5) after taking out the common factor. The word completely in a question means you must keep factorising until no bracket can be broken down further.

Algebraic fractions

Algebraic fractions follow the same rules as number fractions. To simplify, factorise the numerator and denominator fully and cancel common factors, never single terms. For example, (x² − 4)/(x² + 5x + 6) = (x − 2)(x + 2) / ((x + 2)(x + 3)) = (x − 2)/(x + 3). To add or subtract, use a common denominator: 2/(x + 3) − 1/(x − 1) = (2(x − 1) − (x + 3)) / ((x + 3)(x − 1)) = (x − 5)/((x + 3)(x − 1)). Notice the bracket around (x + 3) in the numerator, which makes the subtraction give −x − 3. To multiply, factorise and cancel before multiplying. To divide, multiply by the reciprocal of the second fraction. Leave denominators factorised unless the question asks otherwise, since that form is easier to check and to cancel.

Changing the subject

To change the subject, undo the operations on the new subject in reverse order, doing the same thing to both sides each time. For T = 2π√(L/g), divide by 2π to get T/(2π) = √(L/g), square both sides to get T²/(4π²) = L/g, then multiply by g, giving L = gT²/(4π²). When the new subject appears in two places, collect those terms on one side and factorise. To make x the subject of a(x − b) = cx, expand to ax − ab = cx, gather the x terms to get ax − cx = ab, factorise to x(a − c) = ab and divide, giving x = ab/(a − c). If the subject is squared, as in A = πr², the next step is a square root, r = √(A/π), and when r is a length you take the positive root.

Worked example

Make x the subject of the formula y = (3x + 2)/(x − 1). Hence find x when y = 5.

  1. Multiply both sides by (x − 1) to clear the fraction: y(x − 1) = 3x + 2.
  2. Expand the bracket: xy − y = 3x + 2.
  3. Collect the x terms on one side and the other terms on the other side: xy − 3x = y + 2.
  4. Factorise out x: x(y − 3) = y + 2.
  5. Divide by (y − 3): x = (y + 2)/(y − 3).
  6. Substitute y = 5: x = 7/2 = 3.5.
  7. Check in the original formula: (3(3.5) + 2)/(3.5 − 1) = 12.5/2.5 = 5, which is correct.

Answer: x = (y + 2)/(y − 3), and x = 3.5 when y = 5

Common mistakes

  • Forgetting the middle term when squaring a bracket. (x + 5)² is x² + 10x + 25, not x² + 25. Write it as (x + 5)(x + 5) if you are unsure.
  • Cancelling terms instead of factors. In (x + 6)/(x + 3) you cannot cancel the x or turn 6/3 into 2. Only whole factors that multiply the entire numerator and the entire denominator can be cancelled.
  • Losing a negative sign when subtracting a fraction. In 3/(x − 2) − 2/(x + 1), the numerator is 3(x + 1) − 2(x − 2) = 3x + 3 − 2x + 4. Writing −4 at the end instead of +4 is a very common slip.
  • Stopping factorisation too early. 2x² − 50 = 2(x² − 25) is not complete, because x² − 25 is a difference of two squares. The full answer is 2(x − 5)(x + 5).
  • Leaving the new subject on both sides when changing the subject. An answer such as x = (y + 2 + 3x)/y is not finished. Collect every term containing x on one side and factorise.
  • Square-rooting term by term. From x² = a² + b² you cannot write x = a + b. The correct step is x = √(a² + b²).

Exam tips

  • After factorising, expand your answer in your head to check that it gives back the original expression. This takes seconds and catches sign errors.
  • When asked to simplify an algebraic fraction, your first move should almost always be to factorise top and bottom. A factor that appears in both is a strong hint that you are on the right track.
  • In change-the-subject questions, write one operation per line. Short lines make it easier for you to spot where a sign went wrong and for a marker to follow your method.

Practice questions

1. Factorise completely 6x² − 15x.

Show answer

3x(2x − 5). The highest common factor of 6x² and 15x is 3x, and 6x² ÷ 3x = 2x while 15x ÷ 3x = 5.

2. Simplify (x² − 9)/(x² + x − 6).

Show answer

(x − 3)/(x − 2). The numerator is a difference of two squares, (x − 3)(x + 3), and the denominator factorises to (x + 3)(x − 2). Cancel the common factor (x + 3).

3. Express 3/(x − 2) − 2/(x + 1) as a single fraction in its simplest form.

Show answer

(x + 7)/((x − 2)(x + 1)). Use the common denominator (x − 2)(x + 1). The numerator is 3(x + 1) − 2(x − 2) = 3x + 3 − 2x + 4 = x + 7.

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