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IGCSE / O Level Mathematics · Topic 2 Algebra and graphs · Chapter 11 · pp. 102–121

Algebraic representation and manipulation

Algebra has two opposite moves: expanding turns a product into a sum, and factorising turns a sum back into a product. This chapter practises both, then uses them to rearrange formulae and to tidy algebraic fractions.

Learning objectives

  1. Expand a single bracket, and expand and simplify expressions with several brackets.
  2. Expand a pair of brackets with the grid method, including squares such as (x − 3)².
  3. Factorise by taking out a common factor, by grouping, and with the difference of two squares.
  4. Factorise quadratic expressions such as x² + 5x + 6 and 3x² + 7x − 6.
  5. Substitute into expressions and rearrange formulae, including formulae with squares and roots.
  6. Simplify algebraic fractions: multiply, divide, add and subtract them.

Book exercises (pp. 102–121): 11.1 to 11.18 and Student assessments 1 to 4. Exercises 11.7 to 11.18 and Student assessments 3 and 4 are Extended content.

Expanding a bracket

🔑 Key ideas

  • To remove a bracket, multiply every term inside by whatever is outside.
  • Watch the signs: a minus outside the bracket changes the sign of every term inside.
  • After expanding, collect like terms (terms with exactly the same letters).
3 ( x + 4 ) = 3x + 12 3 × x = 3x 3 × 4 = 12
The 3 outside the bracket visits both terms inside: 3 × x and 3 × 4.

📘 Example 1 · Expand 5x(2y + 3) and −4p(2p − q + r²)

  1. 5x × 2y = 10xy and 5x × 3 = 15x, so the first is 10xy + 15x.
  2. −4p × 2p = −8p².
  3. −4p × (−q) = +4pq.
  4. −4p × r² = −4pr². The second is −8p² + 4pq − 4pr².

📘 Example 2 · Expand and simplify 3a − 2(2a + 4)

  1. Expand the bracket: −2 × 2a = −4a and −2 × 4 = −8.
  2. Write it out: 3a − 4a − 8.
  3. Collect the a terms: 3a − 4a = −a.
  4. Answer: −a − 8.
  5. Check with a = 1: 3 − 2 × 6 = −9, and −1 − 8 = −9. ✔

Expanding a pair of brackets

🔑 Key ideas

  • Multiply every term in one bracket by every term in the other: 2 × 2 = 4 products.
  • The grid method keeps them tidy: one bracket along the top, the other down the side.
  • The two middle cells are like terms, so they join into one x term.
  • (x + 3)(x − 3) has no middle term: the x terms cancel. That is the difference of two squares.
× x+ 5x+ 3x²5x3x15
(x + 3)(x + 5): x² + 5x + 3x + 15 = x² + 8x + 15. The two shaded middle cells are like terms.

📘 Example 3 · Expand and simplify (x − 4)(x + 6)

  1. Multiply x by each term of the second bracket: x × x = x² and x × 6 = 6x.
  2. Multiply −4 by each term: −4 × x = −4x and −4 × 6 = −24.
  3. Write all four: x² + 6x − 4x − 24.
  4. Collect: 6x − 4x = 2x.
  5. Answer: x² + 2x − 24.
  6. Check with x = 1: (−3)(7) = −21, and 1 + 2 − 24 = −21. ✔

📘 Example 4 · Expand (x − 3)², then (a + b)(a − b)

  1. A square means the bracket twice: (x − 3)(x − 3).
  2. Four products: x² − 3x − 3x + 9.
  3. Collect: x² − 6x + 9. Not x² + 9: the middle term is easy to forget.
  4. (a + b)(a − b) = a² − ab + ab − b².
  5. The middle terms cancel: a² − b².

Further expansion Extended

The same method works when the x terms have numbers in front, and with three brackets (multiply two first, then the third).

📘 Example 5 · Expand (2x − 3)(3x − 6)

  1. 2x × 3x = 6x² and 2x × (−6) = −12x.
  2. −3 × 3x = −9x and −3 × (−6) = +18.
  3. Write them: 6x² − 12x − 9x + 18.
  4. Collect: −12x − 9x = −21x.
  5. Answer: 6x² − 21x + 18.

📘 Example 6 · Expand (x + 1)(x + 2)(x + 3)

  1. First pair: (x + 1)(x + 2) = x² + 3x + 2.
  2. Multiply by (x + 3): x² × (x + 3) = x³ + 3x², and 3x × (x + 3) = 3x² + 9x, and 2 × (x + 3) = 2x + 6.
  3. Write them: x³ + 3x² + 3x² + 9x + 2x + 6.
  4. Collect: x³ + 6x² + 11x + 6.
  5. Check with x = 1: 2 × 3 × 4 = 24, and 1 + 6 + 11 + 6 = 24. ✔

🧪 Tool: expand (ax + b)(cx + d) with the grid method

(x + ) (x + )

Whole numbers from −99 to 99. Negative numbers are fine: type −4 for “− 4”.

× x+ 5x+ 3x²5x3x15
The grid for your brackets: the row term times the column term goes in each cell.

    ✏️ Try it

    1. Expand −3(2x − 5).
    2. Expand and simplify (x + 7)(x − 2).
    3. Expand and simplify (3x + 1)(2x − 4).
    Answers

    1. −6x + 15 · 2. x² − 2x + 7x − 14 = x² + 5x − 14 · 3. 6x² − 12x + 2x − 4 = 6x² − 10x − 4

    Next: factorising →

    Simple factorising

    🔑 Key ideas

    • Factorising is expanding backwards: find the largest common factor of all the terms and put it outside a bracket.
    • Take out numbers and letters: the factor of 6ax − 12ay − 18a² is 6a.
    • Check by expanding: you must get the original back.
    • Always take out a common factor first, before any other method.
    1. Take out the common factor 2 terms difference of squares 3 terms quadratic: find a pair 4 terms grouping in pairs 4a² − 9b² (2a + 3b)(2a − 3b) x² + 5x + 6 (x + 2)(x + 3) 6x + 3 + 2xy + y (3 + y)(2x + 1)
    Choose the method by counting the terms (after taking out any common factor).

    📘 Example 7 · Factorise 8p − 6q + 10r and 6ax − 12ay − 18a²

    1. 8p − 6q + 10r: the numbers 8, 6 and 10 all divide by 2, and there is no common letter.
    2. Divide each term by 2: 4p, −3q, 5r.
    3. First answer: 2(4p − 3q + 5r).
    4. 6ax − 12ay − 18a²: the numbers share 6 and every term has an a, so the factor is 6a.
    5. Divide each term by 6a: x, −2y, −3a.
    6. Second answer: 6a(x − 2y − 3a). Check: 6a × (−3a) = −18a². ✔

    Factorisation by grouping Extended

    With four terms, split them into two pairs. Take a common factor out of each pair. If the brackets that are left match, they become a common factor too.

    📘 Example 8 · Factorise 6x + 3 + 2xy + y

    1. Pair the terms: (6x + 3) and (2xy + y).
    2. First pair: 6x + 3 = 3(2x + 1).
    3. Second pair: 2xy + y = y(2x + 1).
    4. Both give the bracket (2x + 1), so it is a common factor: 3(2x + 1) + y(2x + 1).
    5. Answer: (3 + y)(2x + 1).
    6. Check with x = 1, y = 1: 6 + 3 + 2 + 1 = 12, and (4)(3) = 12. ✔

    💡 If the two brackets do not match, change the order of the four terms or take out a minus sign (as in ax + ay − bx − by = a(x + y) − b(x + y)).

    Difference of two squares Extended

    🔑 Key ideas

    • a² − b² = (a + b)(a − b): it is the expansion of (a + b)(a − b) read backwards.
    • It needs two terms, both perfect squares, with a minus between them.
    • x² + 9 (a plus sign) does not factorise this way.
    • Numbers can be squares too: 25 = 5², 1 = 1², 4x² = (2x)².

    📘 Example 9 · Factorise 4a² − 9b² and p⁴ − q⁴

    1. 4a² = (2a)² and 9b² = (3b)².
    2. Write it as (2a)² − (3b)².
    3. First answer: (2a + 3b)(2a − 3b).
    4. p⁴ = (p²)² and q⁴ = (q²)², so p⁴ − q⁴ = (p² + q²)(p² − q²).
    5. The second bracket is again a difference of two squares: p² − q² = (p + q)(p − q).
    6. Answer: (p² + q²)(p + q)(p − q).

    📘 Example 10 · Evaluate 13² − 7² and 6.25² − 3.75² by factorising

    1. 13² − 7² = (13 + 7)(13 − 7).
    2. = 20 × 6 = 120.
    3. 6.25² − 3.75² = (6.25 + 3.75)(6.25 − 3.75).
    4. = 10 × 2.5 = 25.
    5. Check the first: 169 − 49 = 120. ✔

    ✏️ Try it

    1. Factorise 8x²y − 4xy².
    2. Factorise 25x² − 16.
    3. Evaluate 52² − 48² by factorising.
    Answers

    1. 4xy(2x − y) · 2. (5x + 4)(5x − 4) · 3. (52 + 48)(52 − 48) = 100 × 4 = 400

    Next: factorising quadratics →

    Factorising quadratic expressions Extended

    🔑 Key ideas

    • A quadratic expression has x² as its highest power: x² + 5x + 6, or 3x² + 7x − 6.
    • It factorises into two brackets. For x² + bx + c, find two numbers that multiply to c and add to b.
    • For ax² + bx + c, find two numbers that multiply to a × c and add to b, then split the middle term and group.
    • Always check by expanding the brackets again.
    × x+ 2x+ 3x²2x3x6
    Factorising x² + 5x + 6 by the grid: x² and 6 go in the corners, then 3 and 2 (multiply to 6, add to 5) make 3x + 2x = 5x.
    Signs in x² + bx + cThe two numbers areExample
    c positive, b positiveboth positivex² + 5x + 6 = (x + 2)(x + 3)
    c positive, b negativeboth negativex² − 7x + 12 = (x − 3)(x − 4)
    c negativeone of each sign; the bigger one has b’s signx² + 2x − 24 = (x − 4)(x + 6)

    📘 Example 11 · Factorise x² + 2x − 24

    1. Find two numbers that multiply to −24 and add to +2.
    2. c is negative, so one is positive and one negative.
    3. Try pairs of 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6. The difference must be 2: that is 4 and 6.
    4. The bigger one (6) is positive, as b is positive: +6 and −4.
    5. Answer: (x + 6)(x − 4).
    6. Check: x² − 4x + 6x − 24 = x² + 2x − 24. ✔

    📘 Example 12 · Factorise 3x² + 7x − 6

    1. Multiply a × c: 3 × (−6) = −18.
    2. Find two numbers that multiply to −18 and add to +7: they are +9 and −2.
    3. Split the middle term: 3x² + 9x − 2x − 6.
    4. Group: 3x(x + 3) − 2(x + 3).
    5. Take out the common bracket: (3x − 2)(x + 3).
    6. Check: 3x² + 9x − 2x − 6 = 3x² + 7x − 6. ✔

    💡 Perfect squares: x² + 6x + 9 = (x + 3)² and x² − 12x + 36 = (x − 6)². The two numbers are equal. Take out any common factor first: 3x² + 3x − 18 = 3(x² + x − 6) = 3(x + 3)(x − 2).

    🧪 Tool: factorise ax² + bx + c

    x² + x +

    Whole numbers: a from 1 to 30, b and c from −99 to 99. If nothing fits, the expression does not factorise with whole numbers.

    × 3x− 2x+ 33x²−2x9x−6
    The grid for the answer: the corners are ax² and c, the two middle cells hold the split middle term.

      ✏️ Try it

      1. Factorise x² + 7x + 10.
      2. Factorise x² − x − 12.
      3. Factorise 2x² + 5x + 3.
      Answers

      1. 2 × 5 = 10 and 2 + 5 = 7: (x + 2)(x + 5) · 2. −4 × 3 = −12 and −4 + 3 = −1: (x − 4)(x + 3) · 3. 2 × 3 = 6, pair 2 and 3: 2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)

      Next: substitution and formulae →

      Substitution

      🔑 Key ideas

      • Substitute means replace each letter by its value. Do powers first, then × and ÷, then + and −.
      • Put a negative value in brackets: if c = −5 then c² = (−5)² = 25, not −25.
      • Minus a negative is a plus: 9 − (−8) = 17.

      📘 Example 13 · a = 3, b = 4, c = −5. Evaluate a² + b² + c² and 3a(2b − 3c)

      1. a² + b² + c² = 3² + 4² + (−5)².
      2. = 9 + 16 + 25 = 50.
      3. 3a(2b − 3c) = 3 × 3 × (2 × 4 − 3 × (−5)).
      4. Bracket: 8 + 15 = 23.
      5. 9 × 23 = 207.

      Rearranging formulae

      🔑 Key ideas

      • In a = 2b + c, the letter a is the subject. To make b the subject, rearrange so b is alone on one side.
      • Do the same thing to both sides, undoing the operations in reverse order.
      • If the subject appears twice, collect those terms on one side and factorise it out.
      • To undo a square, take the square root. It gives ±, but a length uses the positive root only.
      Clearfractionsand roots Expandbrackets Subject termsto one side Factoriseout thesubject Divide bythe bracket Use only the steps you need, in this order.
      A route for rearranging any formula. Most formulae need only two or three of these steps.

      📘 Example 14 · Make y the subject of 3x − 7y = 4z

      1. Subtract 3x from both sides: −7y = 4z − 3x.
      2. Divide both sides by −7: y = (4z − 3x) ÷ (−7).
      3. Change both signs on top to remove the minus below: y = (3x − 4z)/7.
      4. Check with x = 5, z = 2: 15 − 7y = 8 gives y = 1, and (15 − 8)/7 = 1. ✔

      📘 Example 15 · Make l the subject of T = 2π√(l/g)

      1. Divide both sides by 2π: T/(2π) = √(l/g).
      2. Square both sides: T²/(4π²) = l/g.
      3. Multiply both sides by g: l = gT²/(4π²).
      4. Check: if T = 2π then l = g × 4π²/(4π²) = g, and T = 2π√(g/g) = 2π. ✔

      📘 Example 16 · Make y the subject of x(w + y) = 2y Extended

      1. Expand: xw + xy = 2y.
      2. y appears twice, so collect the y terms on one side: xw = 2y − xy.
      3. Factorise out y: xw = y(2 − x).
      4. Divide by (2 − x): y = xw/(2 − x).
      5. Check with x = 1, w = 3: y = 3/1 = 3, and 1 × (3 + 3) = 6 = 2 × 3. ✔

      Using formulae Extended

      📘 Example 17 · V = πr²h. Find r when V = 6000 cm³ and h = 75 cm

      1. Make r the subject: divide by πh to get r² = V/(πh).
      2. Take the square root: r = √(V/(πh)).
      3. Substitute: r = √(6000/(π × 75)) = √(80/π).
      4. 80 ÷ π = 25.46, and √25.46 = 5.046.
      5. Answer: r = 5.05 cm (3 s.f.).
      6. Check: π × 5.05² × 75 = π × 25.5 × 75 ≈ 6010. ✔ (close, because of rounding)

      📘 Example 18 · C = (5/9)(F − 32). Convert 150 °F, then make F the subject

      1. Substitute F = 150: C = (5/9)(150 − 32) = (5/9) × 118.
      2. = 590 ÷ 9 = 65.6 °C (1 d.p.).
      3. Multiply both sides by 9/5: 9C/5 = F − 32.
      4. Add 32: F = 9C/5 + 32.
      5. Check with C = 160: 9 × 160/5 + 32 = 288 + 32 = 320 °F, and 5/9 × (320 − 32) = 5/9 × 288 = 160. ✔

      ✏️ Try it

      1. p = 4, q = −2, r = 3. Evaluate q² − r².
      2. With the same values, evaluate p(q − r).
      3. Make t the subject of v = u + at.
      Answers

      1. (−2)² − 3² = 4 − 9 = −5 · 2. 4 × (−2 − 3) = 4 × (−5) = −20 · 3. v − u = at, so t = (v − u)/a

      Next: algebraic fractions →

      Algebraic fractions Extended

      🔑 Key ideas

      • The rules are exactly the rules for number fractions, with letters.
      • Multiply: tops together, bottoms together, then cancel. Divide: flip the second fraction and multiply.
      • Add or subtract: the bottoms must match first. Use the lowest common multiple.
      • Cancel only factors (things multiplied), never terms (things added): (x + 3)/3 is not x.
      Factoriseevery bottom Find thelowest commonmultiple Multiply topand bottom tomatch it Add orsubtractthe tops Factorise thetop and cancel Adding or subtracting: follow all five boxes. Multiplying: just the last one.
      The route for adding and subtracting algebraic fractions.

      📘 Example 19 · Simplify 12pq²/(3p), then (4/x) ÷ (2/x²)

      1. 12pq²/(3p): 12 ÷ 3 = 4.
      2. The p on top cancels the p below.
      3. First answer: 4q².
      4. (4/x) ÷ (2/x²): flip the second fraction and multiply: 4/x × x²/2.
      5. Multiply: 4x²/(2x). Cancel 2 and one x.
      6. Second answer: 2x.

      📘 Example 20 · Simplify x/3 + x/4 and 2/a + 3/b

      1. x/3 + x/4: the lowest common multiple of 3 and 4 is 12.
      2. x/3 = 4x/12 and x/4 = 3x/12.
      3. Add the tops: 4x + 3x = 7x. First answer: 7x/12.
      4. 2/a + 3/b: the common bottom is ab.
      5. 2/a = 2b/(ab) and 3/b = 3a/(ab).
      6. Second answer: (2b + 3a)/(ab).
      7. Check with a = 1, b = 2: 2 + 1.5 = 3.5, and (4 + 3)/2 = 3.5. ✔

      📘 Example 21 · Simplify 5/(3a) − 1/(2a)

      1. The lowest common multiple of 3a and 2a is 6a.
      2. 5/(3a) = 10/(6a) (top and bottom × 2).
      3. 1/(2a) = 3/(6a) (top and bottom × 3).
      4. Subtract the tops: 10 − 3 = 7.
      5. Answer: 7/(6a).
      6. Check with a = 1: 5/3 − 1/2 = 10/6 − 3/6 = 7/6. ✔

      📘 Example 22 · Simplify 2/(x + 1) + 3/(x + 2) and 5/(p + 3) − 3/(p − 5)

      1. The bottoms have no common factor, so the common bottom is their product (x + 1)(x + 2).
      2. Multiply each top by the missing bracket: 2(x + 2) + 3(x + 1) over (x + 1)(x + 2).
      3. Expand the top: 2x + 4 + 3x + 3 = 5x + 7.
      4. First answer: (5x + 7)/((x + 1)(x + 2)).
      5. Second: the common bottom is (p + 3)(p − 5), with top 5(p − 5) − 3(p + 3).
      6. Expand with care over the minus: 5p − 25 − 3p − 9 = 2p − 34.
      7. Second answer: (2p − 34)/((p + 3)(p − 5)).
      8. Check the first with x = 1: 2/2 + 3/3 = 2, and 12/(2 × 3) = 2. ✔

      📘 Example 23 · Simplify (x² − 4)/(x² + 4x + 4)

      1. Factorise the top (difference of two squares): (x − 2)(x + 2).
      2. Factorise the bottom (perfect square): (x + 2)(x + 2).
      3. Cancel the common factor (x + 2).
      4. Answer: (x − 2)/(x + 2).
      5. Check with x = 1: (1 − 4)/(1 + 4 + 4) = −3/9 = −1/3, and (1 − 2)/(1 + 2) = −1/3. ✔

      ✏️ Try it

      1. Simplify 18a²b/(6ab²).
      2. Simplify 1/(2x) + 1/(3x).
      3. Simplify (x² − 1)/(x + 1).
      Answers

      1. 18 ÷ 6 = 3, then a²/a = a and b/b² = 1/b: 3a/b · 2. 3/(6x) + 2/(6x) = 5/(6x) · 3. (x + 1)(x − 1)/(x + 1) = x − 1

      Next: summary and quiz →

      Chapter summary

      • Expanding: multiply every term in one bracket by every term in the other, then collect like terms. (a + b)(a − b) = a² − b².
      • (x ± k)² has a middle term: (x − 3)² = x² − 6x + 9.
      • Factorising is expanding backwards. Common factor first; then count the terms: 2 (difference of two squares), 3 (quadratic), 4 (grouping).
      • x² + bx + c: find two numbers that multiply to c and add to b. For ax² + bx + c, use a × c, split the middle term, then group.
      • Substitute negative values in brackets: (−5)² = 25.
      • Rearrange by doing the same to both sides. Squares are undone by square roots. If the subject appears twice, factorise it out.
      • Algebraic fractions: cancel factors only. To add or subtract, use the lowest common multiple of the bottoms.

      📄 Practice questions

      The book’s Exercises 11.1 to 11.18 and Student assessments 1 to 4 are in the printable assignment for this chapter.

      Mini-quiz

      1. Expand 3(2x − 5).
      2. Expand and simplify 4x − 2(x − 3).
      3. Expand and simplify (x + 4)(x − 2).
      4. Expand (x − 5)².
      5. Factorise fully 12ab − 8a².
      6. Factorise 9x² − 25.
      7. Factorise x² − 3x − 10.
      8. By factorising, work out 98² − 2².
      9. Find a² − b³ when a = 3 and b = −2.
      10. Make x the subject of y = 3x − 5.
      11. Make r the subject of A = πr², where r > 0.
      12. Simplify x/3 + x/4.

      Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 11 (Algebraic representation and manipulation). Explanations, the Try it answers, the tools and the quiz are Guru Jazzy’s own. This page is not made or endorsed by Cambridge or Hodder Education.