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Three binomials and non-monic factorisation

3 exam-style questions. Answers and the mark for each step are on the last page.

  1. Question 1Non-calculator · 4 marks

    Factorise each expression.

    (a) Factorise 2x2+7x+32x^2 + 7x + 3 (2)

    (b) Factorise 6x2−x−26x^2 - x - 2 (2)

  2. Question 2Non-calculator · 4 marks

    Consider (2x+3)(x−4)(2x+3)(x-4).

    (a) Expand and simplify. (2)

    (b) Now multiply your result by (x + 1) and expand. (2)

  3. Question 3Non-calculator · 4 marks

    Consider 6x2+38x+566x^2+38x+56.

    (a) Factorise fully. (2)

    (b) Solve the equation obtained by setting the expression equal to zero. (2)

Answers and marks

Question 1

(a) (2x+1)(x+3)(2x + 1)(x + 3)

  • M1 Splitting to 2x2+6x+x+32x^2 + 6x + x + 3, or brackets (2x±a)(x±b)(2x \pm a)(x \pm b) with ab=3ab = 3.
  • A1 (2x+1)(x+3)(2x + 1)(x + 3).

(b) (3x−2)(2x+1)(3x - 2)(2x + 1)

  • M1 Splitting the middle term as −4x+3x-4x + 3x, or a pair of brackets with the right x2x^2 and constant terms.
  • A1 (3x−2)(2x+1)(3x - 2)(2x + 1).

Question 2

(a) 2x2−5x−122x^{2}-5x-12

  • M1 Multiply all four term pairs and collect the middle terms.
  • A1 Correct answer: 2x2−5x−122x^{2}-5x-12

(b) 2x3−3x2−17x−122x^{3}-3x^{2}-17x-12

  • M1 Distribute both x and 1 across all three terms.
  • A1 Correct answer: 2x3−3x2−17x−122x^{3}-3x^{2}-17x-12

Question 3

(a) 2(x+4)(3x+7)2(x+4)(3x+7)

  • M1 Split the middle term: 38x = 14x + 24x, since 14 × 24 = 6 × 56.
  • A1 Correct answer: 2(x+4)(3x+7)2(x+4)(3x+7)

(b) −4,−7/3-4, -7/3

  • M1 Set each non-monic factor equal to zero.
  • A1 Correct answer: −4,−7/3-4, -7/3

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Three binomials and non-monic factorisation

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Independent practice for Pearson Edexcel GCSE Mathematics (1MA1), not endorsed by Pearson.

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