By /
IGCSE Maths · Edexcel 4XMA1 · Unit 1 Higher · Algebraic manipulation

Expanding Triple Brackets — Worked Solutions

Full worked solutions to all 12 questions (36 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 4–5)
  1. (2 marks)Expand and simplify (x + 1)(x + 2)(x + 5).
    Solution
    1. (x + 1)(x + 2) = x² + 3x + 2.
    2. (x² + 3x + 2)(x + 5) = x³ + 5x² + 3x² + 15x + 2x + 10.
    x³ + 8x² + 17x + 10
  2. (3 marks)Expand and simplify (x + 4)(x − 2)(x + 1).
    Solution
    1. (x + 4)(x − 2) = x² + 2x − 8.
    2. (x² + 2x − 8)(x + 1) = x³ + x² + 2x² + 2x − 8x − 8.
    x³ + 3x² − 6x − 8
  3. (3 marks)Expand and simplify (x − 3)(x − 1)(x + 5).
    Solution
    1. (x − 3)(x − 1) = x² − 4x + 3.
    2. (x² − 4x + 3)(x + 5) = x³ + 5x² − 4x² − 20x + 3x + 15.
    x³ + x² − 17x + 15
  4. (3 marks)Expand and simplify (x + 2)³.
    Solution
    1. (x + 2)² = x² + 4x + 4.
    2. (x² + 4x + 4)(x + 2) = x³ + 2x² + 4x² + 8x + 4x + 8.
    x³ + 6x² + 12x + 8
Section B — Core (grades 5–7)
  1. (3 marks)Expand and simplify (2x + 3)(x − 1)(x + 4).
    Solution
    1. (2x + 3)(x − 1) = 2x² + x − 3.
    2. (2x² + x − 3)(x + 4) = 2x³ + 8x² + x² + 4x − 3x − 12.
    2x³ + 9x² + x − 12
  2. (3 marks)Expand and simplify (3x − 2)(2x + 1)(x − 4).
    Solution
    1. (3x − 2)(2x + 1) = 6x² − x − 2.
    2. (6x² − x − 2)(x − 4) = 6x³ − 24x² − x² + 4x − 2x + 8.
    6x³ − 25x² + 2x + 8
  3. (3 marks)Expand and simplify (2x − 5)³.
    Solution
    1. (2x − 5)² = 4x² − 20x + 25.
    2. (4x² − 20x + 25)(2x − 5) = 8x³ − 20x² − 40x² + 100x + 50x − 125.
    8x³ − 60x² + 150x − 125
  4. (3 marks)Show that (x + 3)(x − 3)(2x + 1) = 2x³ + x² − 18x − 9.
    Solution
    1. (x + 3)(x − 3) = x² − 9 (difference of two squares).
    2. (x² − 9)(2x + 1) = 2x³ + x² − 18x − 9, as required.
    Shown: 2x³ + x² − 18x − 9
  5. (3 marks)A cuboid has length (x + 1) cm, width (x + 2) cm and height (2x − 3) cm. Show that its volume, in cm³, is 2x³ + 3x² − 5x − 6.
    Solution
    1. V = (x + 1)(x + 2)(2x − 3).
    2. (x + 1)(x + 2) = x² + 3x + 2.
    3. (x² + 3x + 2)(2x − 3) = 2x³ − 3x² + 6x² − 9x + 4x − 6 = 2x³ + 3x² − 5x − 6.
    V = 2x³ + 3x² − 5x − 6 cm³
Section C — Stretch (grades 8–9)
  1. (2 marks)Find the coefficient of x² in the expansion of (x + 2)(x − 5)(3x + 1), without expanding fully.
    Solution
    1. x² terms come from (x)(x)(1) + (x)(−5)(3x) + (2)(x)(3x).
    2. = x² − 15x² + 6x² = −8x².
    −8
  2. (4 marks)(x + a)(x − 4)(x + 1) ≡ x³ + bx² − 7x + c, where a, b and c are integers. Find a, b and c.
    Solution
    1. (x − 4)(x + 1) = x² − 3x − 4.
    2. (x + a)(x² − 3x − 4) = x³ + (a − 3)x² + (−4 − 3a)x − 4a.
    3. Compare x terms: −4 − 3a = −7, so a = 1.
    4. b = a − 3 = −2 and c = −4a = −4.
    a = 1, b = −2, c = −4
  3. (4 marks)(a) Show that (n + 1)³ − n³ = 3n² + 3n + 1.
    (b) Hence prove that the difference between the cubes of two consecutive integers is always odd.
    Solution
    1. (a) (n + 1)³ = n³ + 3n² + 3n + 1, so (n + 1)³ − n³ = 3n² + 3n + 1.
    2. (b) 3n² + 3n + 1 = 3n(n + 1) + 1.
    3. n(n + 1) is the product of two consecutive integers, so one of them is even and n(n + 1) is even.
    4. So 3n(n + 1) is even, and 3n(n + 1) + 1 is odd.
    Proved: 3n(n + 1) is even, so the difference is odd
← Back to the Edexcel Unit 1 resource listThe Vector Academy · thevectoracademy.in

Book a free trial class

Tell us a little about your requirement.