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Exam questions · Maths · Algebra

Factorising Expressions

  • 7 exam questions
  • 16 marks
  • 10 quick checks
  1. 1 Factorise [2 marks]

    Factorise \(6x + 15\).

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    Model answer

    The highest common factor of 6 and 15 is 3, so \(6x + 15 = 3(2x + 5)\).

    Mark scheme

    • \(3(\ldots)\) or a common factor taken out correctly — M1
    • \(3(2x + 5)\) — A1
  2. 2 Factorise [3 marks]

    Factorise fully (a) \(12a + 18\) (1 mark) (b) \(6x^2y - 9xy^2\) (2 marks)

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    Model answer

    (a) The HCF of 12 and 18 is 6, so \(6(2a + 3)\). (b) The HCF of the numbers is 3 and both terms contain \(x\) and \(y\), so \(3xy(2x - 3y)\).

    Mark scheme

    • (a) \(6(2a + 3)\) — B1
    • (b) A correct partial factorisation, such as \(3(2x^2y - 3xy^2)\) or \(xy(6x - 9y)\) — M1
    • (b) \(3xy(2x - 3y)\) — A1
  3. 3 Factorise [2 marks]

    Factorise \(x^2 + 8x + 15\).

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    Model answer

    The numbers that multiply to give 15 and add to give 8 are 3 and 5, so \(x^2 + 8x + 15 = (x + 3)(x + 5)\).

    Mark scheme

    • \((x + a)(x + b)\) with \(ab = 15\) or \(a + b = 8\) — M1
    • \((x + 3)(x + 5)\) — A1
  4. 4 Factorise [2 marks]

    Factorise \(x^2 - 2x - 15\).

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    Model answer

    The numbers that multiply to give \(-15\) and add to give \(-2\) are \(-5\) and \(3\), so \(x^2 - 2x - 15 = (x - 5)(x + 3)\).

    Mark scheme

    • \((x + a)(x + b)\) with \(ab = -15\) or \(a + b = -2\) — M1
    • \((x - 5)(x + 3)\) — A1
  5. 5 Factorise [3 marks]

    (a) Factorise \(x^2 - 81\). (1 mark) (b) Factorise fully \(2y^2 - 50\). (2 marks)

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    Model answer

    (a) This is a difference of two squares: \(x^2 - 9^2 = (x + 9)(x - 9)\). (b) Take out the common factor 2 first: \(2y^2 - 50 = 2(y^2 - 25) = 2(y + 5)(y - 5)\).

    Mark scheme

    • (a) \((x + 9)(x - 9)\) — B1
    • (b) \(2(y^2 - 25)\) — M1
    • (b) \(2(y + 5)(y - 5)\) — A1
  6. 6 Factorise [2 marks]

    Factorise \(3x^2 - 10x - 8\).

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    Model answer

    \(3 \times (-8) = -24\), and \(-12\) and \(2\) multiply to \(-24\) and add to \(-10\). So \(3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)\).

    Mark scheme

    • \((3x + a)(x + b)\) with \(ab = -8\) or the middle term correctly split as \(-12x + 2x\) — M1
    • \((3x + 2)(x - 4)\) — A1
  7. 7 Work out [2 marks]

    Work out the value of \(101^2 - 99^2\). You must show your working and you must not use a calculator.

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    Model answer

    Using the difference of two squares, \(101^2 - 99^2 = (101 + 99)(101 - 99) = 200 \times 2 = 400\).

    Mark scheme

    • \((101 + 99)(101 - 99)\) — M1
    • 400 — A1

Quick check

  1. 1

    Factorise \(x^2 - 10x + 25\).

    1. A\((x - 5)^2\)
    2. B\((x + 5)^2\)
    3. C\((x - 5)(x + 5)\)
    4. D\((x - 25)(x + 1)\)
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    A: \((x - 5)^2\)

    The numbers are \(-5\) and \(-5\), so this is the perfect square \((x - 5)^2\).

  2. 2

    Simplify \(\dfrac{x^2 - 9}{x + 3}\).

    1. A\(x - 3\)
    2. B\(x^2 - 3\)
    3. C\(x + 3\)
    4. D\(-3\)
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    A: \(x - 3\)

    \(x^2 - 9 = (x + 3)(x - 3)\), so the \((x + 3)\) cancels, leaving \(x - 3\).

  3. 3

    Factorise \(6x + 15\).

    1. A\(6(x + 15)\)
    2. B\(3(x + 5)\)
    3. C\(6x(1 + 15)\)
    4. D\(3(2x + 5)\)
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    D: \(3(2x + 5)\)

    The highest common factor of 6 and 15 is 3, and \(3(2x + 5) = 6x + 15\).

  4. 4

    Factorise fully \(8x^2 - 12x\).

    1. A\(4x(2x - 12)\)
    2. B\(4(2x^2 - 3x)\)
    3. C\(4x(2x - 3)\)
    4. D\(2x(4x - 6)\)
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    C: \(4x(2x - 3)\)

    The HCF of 8 and 12 is 4, and \(x\) is in both terms, so \(4x\) comes out: \(4x(2x - 3)\).

  5. 5

    Which expression expands to \(x^2 + x - 12\)?

    1. A\((x + 4)(x - 3)\)
    2. B\((x + 6)(x - 2)\)
    3. C\((x - 4)(x + 3)\)
    4. D\((x + 12)(x - 1)\)
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    A: \((x + 4)(x - 3)\)

    \((x + 4)(x - 3) = x^2 - 3x + 4x - 12 = x^2 + x - 12\).

  6. 6

    Factorise \(x^2 + 9x + 18\).

    1. A\((x - 3)(x - 6)\)
    2. B\((x + 3)(x + 6)\)
    3. C\((x + 2)(x + 9)\)
    4. D\((x + 1)(x + 18)\)
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    B: \((x + 3)(x + 6)\)

    The numbers that multiply to 18 and add to 9 are 3 and 6.

  7. 7

    Factorise \(x^2 - 36\).

    1. A\(x(x - 36)\)
    2. B\((x + 6)(x - 6)\)
    3. C\((x - 6)^2\)
    4. D\((x - 18)(x + 2)\)
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    B: \((x + 6)(x - 6)\)

    This is the difference of two squares: \(x^2 - 6^2 = (x + 6)(x - 6)\).

  8. 8

    Which of these cannot be factorised as a difference of two squares?

    1. A\(9 - y^2\)
    2. B\(x^2 + 16\)
    3. C\(4x^2 - 9\)
    4. D\(x^2 - 25\)
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    B: \(x^2 + 16\)

    A difference of two squares needs a minus sign. \(x^2 + 16\) is a sum of squares, so it does not factorise this way.

  9. 9

    Using \(a^2 - b^2 = (a + b)(a - b)\), work out \(52^2 - 48^2\).

    1. A4
    2. B400
    3. C100
    4. D40
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    B: 400

    \((52 + 48)(52 - 48) = 100 \times 4 = 400\).

  10. 10

    Factorise \(2x^2 + 7x + 3\).

    1. A\((x + 7)(2x + 3)\)
    2. B\((2x + 1)(x + 3)\)
    3. C\((2x + 3)(x + 1)\)
    4. D\((2x - 1)(x - 3)\)
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    B: \((2x + 1)(x + 3)\)

    \(2 \times 3 = 6\), and 6 and 1 add to 7, so \(2x^2 + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3)\).