[discia] GCSE Maths › Algebra › Expanding and Factorising Quadratics › Practise
  1. Non-calculator 1 mark
    Which one of these expressions is equivalent to \(x^2 + 7x + 12\)?
    A \((x + 12)(x + 1)\) B \((x + 3)(x + 4)\) C \((x - 3)(x - 4)\) D \((x + 2)(x + 5)\)
    Write down the letter of your answer.
    Show answer
    Option B

    Find two numbers that multiply to give \(12\) and add to give \(7\): they are \(3\) and \(4\).

    \(x^2 + 7x + 12 = (x + 3)(x + 4)\), which is option B.

    Check the others by expanding.

    Option A, \((x + 12)(x + 1)\), gives \(x^2 + 13x + 12\). The product is right but the numbers add to \(13\), not \(7\).

    Option C, \((x - 3)(x - 4)\), gives \(x^2 - 7x + 12\). The signs of both numbers are the wrong way round, so the middle term has the wrong sign.

    Option D, \((x + 2)(x + 5)\), gives \(x^2 + 7x + 10\). The numbers add to \(7\) but multiply to \(10\), not \(12\).

  2. Non-calculator 2 marks

    Expand and simplify \((3x - 2)(4x - 5)\)

    Show answer
    \(12x^2 - 23x + 10\)

    Multiply each term in the first bracket by each term in the second.

    \(3x \times 4x = 12x^2\)

    \(3x \times (-5) = -15x\)

    \((-2) \times 4x = -8x\)

    \((-2) \times (-5) = 10\)

    Collect the two \(x\) terms: \(-15x - 8x = -23x\)

    \(12x^2 - 23x + 10\)

    • M1at least three of the four products correct, e.g. \(12x^2\), \(-15x\), \(-8x\), \(10\)
    • A1fully simplified \(12x^2 - 23x + 10\)
  3. Non-calculator 2 marks

    Multiply out the brackets and simplify \((3t - 4)^2\)

    Show answer
    \(9t^2 - 24t + 16\)

    Squaring a bracket means multiplying it by itself.

    \((3t - 4)(3t - 4)\)

    \(3t \times 3t = 9t^2\)

    There are two cross terms, each \(3t \times (-4) = -12t\), so together they give \(-24t\).

    \((-4) \times (-4) = 16\)

    \(9t^2 - 24t + 16\)

    • M1at least three correct terms from \(9t^2\), \(-12t\), \(-12t\), \(16\)
    • A1fully simplified \(9t^2 - 24t + 16\)
  4. Non-calculator 2 marks

    Factorise \(w^2 - 4w - 32\)

    Show answer
    \((w + 4)(w - 8)\)

    Look for two numbers that multiply to give \(-32\) and add to give \(-4\).

    \(4 \times (-8) = -32\) and \(4 - 8 = -4\)

    Those two numbers go straight into the brackets.

    \(w^2 - 4w - 32 = (w + 4)(w - 8)\)

    • M1two numbers with product \(-32\) or sum \(-4\), or one correct bracket \((w + 4)\) or \((w - 8)\)
    • A1fully factorised \((w + 4)(w - 8)\)
  5. Non-calculator 3 marks
    (a)

    Multiply out the brackets and simplify \((k + 9)(k - 9)\)

    (b)

    Factorise completely \(25k^2 - 4\)

    Show answer
    (a) \(k^2 - 81\)
    (b) \((5k + 2)(5k - 2)\)

    Multiplying out gives \(k^2 - 9k + 9k - 81\).

    The two \(k\) terms cancel, leaving

    \(k^2 - 81\)

    Both terms are squares: \(25k^2 = (5k)^2\) and \(4 = 2^2\).

    A difference of two squares factorises as \(A^2 - B^2 = (A + B)(A - B)\), here with \(A = 5k\) and \(B = 2\).

    \(25k^2 - 4 = (5k + 2)(5k - 2)\)

    • (b) M1recognises the squares, e.g. \((5k)^2 - 2^2\), or one correct bracket
    • (b) A1fully factorised \((5k + 2)(5k - 2)\)
  6. Non-calculator 2 marks

    \((k - 5)(k + a) = k^2 - 13k + 40\)

    Calculate the value of \(a\).

    Show answer
    \(-8\)

    The two numbers in the brackets multiply to give the constant term and add to give the \(k\) coefficient. Either one finds \(a\).

    Using the constant term: \((-5) \times a = 40\), so \(a = -8\)

    Checking against the \(k\) term: \(-5 - 8 = -13\), which matches.

    • M1uses the constant term \((-5) \times a = 40\) or the sum \(-5 + a = -13\)
    • A1\(a = -8\)
  7. Non-calculator 2 marks

    Write \(2x^2 - 13x + 20\) as a product of two brackets.

    Show answer
    \((2x - 5)(x - 4)\)

    Multiply the coefficient of \(x^2\) by the constant term: \(2 \times 20 = 40\).

    Find two numbers that multiply to give \(40\) and add to give \(-13\). They are \(-5\) and \(-8\).

    Use them to split the middle term.

    \(2x^2 - 5x - 8x + 20\)

    Factorise the first two terms and the last two terms.

    \(x(2x - 5) - 4(2x - 5)\)

    Both parts share the bracket \((2x - 5)\), so

    \((2x - 5)(x - 4)\)

    • M1splits the middle term as \(- 5x - 8x\), or one correct bracket
    • A1fully factorised \((2x - 5)(x - 4)\)
  8. Non-calculator 4 marks

    You must show your working.

    Show that \((3x + 2)(x + 6) + 2x(x + 5) - 5(x^2 + 6x + 4)\) simplifies to an integer.

    Show answer
    The \(x^2\) terms and the \(x\) terms both cancel, leaving the integer \(-8\).

    Expand each of the three terms on its own.

    The pair of brackets: \((3x + 2)(x + 6) = 3x^2 + 18x + 2x + 12 = 3x^2 + 20x + 12\)

    The term with \(x\) outside its bracket: \(2x(x + 5) = 2x^2 + 10x\)

    The term with the multiplier outside its bracket: \(5(x^2 + 6x + 4) = 5x^2 + 30x + 20\)

    Put the expansions back with their signs. The minus sign in front of the term with the multiplier outside its bracket changes the sign of every term inside it, so that term goes in a bracket:

    \((3x^2 + 20x + 12) + (2x^2 + 10x) - (5x^2 + 30x + 20)\)

    Remove the brackets:

    \(3x^2 + 20x + 12 + 2x^2 + 10x - 5x^2 - 30x - 20\)

    Collect the \(x^2\) terms: \((3 + 2 - 5)x^2 = 0\)

    Collect the \(x\) terms: \((20 + 10 - 30)x = 0\)

    Collect the numbers: \(12 - 20 = -8\)

    The \(x^2\) terms and the \(x\) terms cancel, so the expression simplifies to the integer \(-8\).

    • M1expands the pair of brackets to \(3x^2 + 20x + 12\)
    • M1expands the term with \(x\) outside its bracket to \(2x^2 + 10x\)
    • M1expands the trinomial bracket to \(5x^2 + 30x + 20\)
    • A1applies the subtraction to every term of the term with the multiplier outside its bracket, shows the \(x^2\) and \(x\) terms cancel and states the integer -8
  9. Non-calculator 3 marks

    The diagram shows a sketch of a quadratic curve. It is not drawn accurately.

    xy(1, 0)(5, 0)(0, -10)

    The curve crosses the \(x\)-axis at two of the labelled points and the \(y\)-axis at the third.

    Work out an equation of the curve. Give your answer in the form \(y = ax^2 + bx + c\).

    Show answer
    \(y = -2x^2 + 12x - 10\)

    The curve crosses the \(x\)-axis where \(x = 1\) and \(x = 5\), so

    \(y = a(x - 1)(x - 5)\)

    The curve passes through \((0, -10)\), so put \(x = 0\) and \(y = -10\):

    \(-10 = a(0 - 1)(0 - 5) = 5a\)

    \(a = \frac{ -10 }{ 5 } = -2\)

    So \(y = -2(x - 1)(x - 5)\). Expand the brackets:

    \((x - 1)(x - 5) = x^2 - 6x + 5\)

    \(y = -2x^2 + 12x - 10\)

    • M1factorised form from the roots, \(y = a(x - 1)(x - 5)\)
    • M1uses \((0, -10)\) to find the multiplier, \(a = -2\)
    • A1expands correctly, \(y = -2x^2 + 12x - 10\)
  10. Non-calculator 3 marks

    \((x + a)(x + 4a) \equiv x^2 + bx + 256\)

    Work out the two possible values of \(b\).

    Show answer
    \(b = -40\) or \(b = 40\)

    Expand the brackets:

    \((x + a)(x + 4a) = x^2 + a x + 4a x + 4a^2 = x^2 + 5a x + 4a^2\)

    Compare the numbers term by term: \(b = 5a\) and \(4a^2 = 256\).

    So \(a^2 = 64\), which means \(a = 8\) or \(a = -8\). Both roots must be used.

    When \(a = 8\): \(b = 5 \times 8 = 40\)

    When \(a = -8\): \(b = 5 \times (-8) = -40\)

    • M1expands to get \(x^2 + 5a x + 4a^2\), so \(4a^2 = 256\)
    • M1\(a^2 = 64\), so \(a = 8\) or \(a = -8\)
    • A1both values of b: -40 and 40 (the positive root alone earns no mark)