Hard GCSE Factorising ax2 + bx + c Questions

Challenging, exam-style GCSE Factorising ax2 + bx + c questions with worked solutions. Stretch yourself on the hardest factorising quadratics, common factor, AC method, difference of two squares problems.

factorising quadraticscommon factorAC methoddifference of two squaresfactorisingsolving quadratics
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
(a) Factorise 4x2+12x+94x^2 + 12x + 9. (b) Hence solve 4x2+12x+9=04x^2 + 12x + 9 = 0.
Show worked solution

Worked solution

  1. Check for a perfect square

    4x2+12x+94x^2 + 12x + 9

    The first and last terms are perfect squares, so test for a perfect-square trinomial.

  2. Square root the first term

    4x2=2x\sqrt{4x^2} = 2x

    The first term of the bracket is 2x2x.

  3. Square root the last term

    9=3\sqrt{9} = 3

    The last term of the bracket is 33.

  4. Propose the square

    (2x+3)2(2x + 3)^2

    A perfect square would be (2x+3)2(2x + 3)^2.

  5. Check the middle term

    2×2x×3=12x2 \times 2x \times 3 = 12x

    The middle term matches, so the factorisation is exact.

  6. State the factorisation

    (2x+3)2=(2x+3)(2x+3)(2x + 3)^2 = (2x + 3)(2x + 3)

    Part (a): the quadratic is a perfect square.

  7. Set up the equation

    4x2+12x+9=04x^2 + 12x + 9 = 0

    Part (b): now solve the equation.

  8. Use the factorised form

    (2x+3)2=0(2x + 3)^2 = 0

    Replace the quadratic with its factorised form.

  9. Take the factor to zero

    2x+3=02x + 3 = 0

    A squared bracket is zero only when the bracket itself is zero.

  10. Rearrange

    2x=−32x = -3

    Subtract 33 from both sides.

  11. Solve for xx

    x=−32x = -\tfrac{3}{2}

    Divide both sides by 22.

  12. Note the repeated root

    one (repeated) solution\text{one (repeated) solution}

    Because the bracket is squared, there is a single repeated root.

  13. Interpret the graph

    the curve touches the x-axis\text{the curve touches the } x\text{-axis}

    A repeated root means the parabola just touches the xx-axis at x=−32x = -\tfrac{3}{2}.

  14. Verify by substitution

    4(−32)2+12(−32)+9=9−18+9=04(-\tfrac{3}{2})^2 + 12(-\tfrac{3}{2}) + 9 = 9 - 18 + 9 = 0

    Substituting the root gives zero, as required.

  15. State the answer

    (2x+3)2;x=−32(2x + 3)^2; \quad x = -\tfrac{3}{2}

    The factorisation and the repeated solution.

Answer
(2x+3)2, x=−32(2x + 3)^2, \ x = -\frac{3}{2}
Question 2
6 markschallenging
Solve 6x2−13x+6=06x^2 - 13x + 6 = 0.
Show worked solution

Worked solution

  1. Plan the method

    6x2−13x+6=06x^2 - 13x + 6 = 0

    Factorise the quadratic, then use the null factor law.

  2. Identify the coefficients

    a=6,b=−13,c=6a = 6, \quad b = -13, \quad c = 6

    Read off aa, bb and cc.

  3. Find the product acac

    6×6=366 \times 6 = 36

    Start the AC method.

  4. List factor pairs

    36=1×36,2×18,3×12,4×9,6×636 = 1\times36, 2\times18, 3\times12, 4\times9, 6\times6

    These are the factor pairs of 3636.

  5. Choose the correct pair

    −4+−9=−13,−4×−9=36-4 + -9 = -13, \quad -4 \times -9 = 36

    The pair −4-4 and −9-9 fits both conditions.

  6. Split the middle term

    6x2−4x−9x+66x^2 - 4x - 9x + 6

    Rewrite −13x-13x as -4x - 9x.

  7. Group into pairs

    (6x2−4x)+(−9x+6)(6x^2 - 4x) + (-9x + 6)

    Bracket the terms in twos.

  8. Factorise the first pair

    2x(3x−2)−9x+62x(3x - 2) - 9x + 6

    Take 2x2x out of the first pair.

  9. Factorise the second pair

    2x(3x−2)−3(3x−2)2x(3x - 2) - 3(3x - 2)

    Take the common factor out of the second pair; both leave (3x−2)(3x - 2).

  10. Write as a product

    (3x−2)(2x−3)=0(3x - 2)(2x - 3) = 0

    Factor out the shared bracket.

  11. Check the factorisation

    (3x−2)(2x−3)=6x2−13x+6(3x - 2)(2x - 3) = 6x^2 - 13x + 6

    Expanding confirms the brackets are correct.

  12. Apply the null factor law

    3x−2=0or2x−3=03x - 2 = 0 \quad \text{or} \quad 2x - 3 = 0

    A product equals zero only when a factor equals zero.

  13. Solve the first bracket

    x=23x = \frac{2}{3}

    Rearrange (3x−2)=0(3x - 2) = 0.

  14. Solve the second bracket

    x=32x = \frac{3}{2}

    Rearrange the other linear equation.

  15. State the solutions

    x=32 or x=23x = \frac{3}{2} \text{ or } x = \frac{2}{3}

    The solutions are x=32x = \frac{3}{2} and x=23x = \frac{2}{3}.

Answer
x=32 or x=23x = \frac{3}{2} \text{ or } x = \frac{2}{3}
Question 3
6 markschallenging
The length of a rectangle is 33 cm more than twice its width. The area of the rectangle is 6565 cm2^2. Work out the width of the rectangle.
Show worked solution

Worked solution

  1. Define the variable

    w=width in cmw = \text{width in cm}

    Let the width be ww centimetres.

  2. Write the length

    length=2w+3\text{length} = 2w + 3

    The length is three more than twice the width.

  3. Form the area equation

    w(2w+3)=65w(2w + 3) = 65

    Area equals length times width.

  4. Expand the brackets

    2w2+3w=652w^2 + 3w = 65

    Multiply out the left-hand side.

  5. Rearrange to equal zero

    2w2+3w−65=02w^2 + 3w - 65 = 0

    Move everything to one side to form a quadratic equation.

  6. Identify the coefficients

    a=2,b=3,c=−65a = 2, \quad b = 3, \quad c = -65

    Read off aa, bb and cc.

  7. Find the product acac

    2×(−65)=−1302 \times (-65) = -130

    Apply the AC method.

  8. Choose the pair

    13+(−10)=3,13×(−10)=−13013 + (-10) = 3, \quad 13 \times (-10) = -130

    The numbers 1313 and −10-10 work.

  9. Split the middle term

    2w2+13w−10w−65=02w^2 + 13w - 10w - 65 = 0

    Rewrite 3w3w as 13w−10w13w - 10w.

  10. Group into pairs

    (2w2+13w)−(10w+65)=0(2w^2 + 13w) - (10w + 65) = 0

    Bracket the terms in twos.

  11. Factorise in pairs

    w(2w+13)−5(2w+13)=0w(2w + 13) - 5(2w + 13) = 0

    Both pairs leave (2w+13)(2w + 13).

  12. Write as a product

    (2w+13)(w−5)=0(2w + 13)(w - 5) = 0

    Factor out the shared bracket.

  13. Apply the null factor law

    2w+13=0orw−5=02w + 13 = 0 \quad \text{or} \quad w - 5 = 0

    One of the factors must be zero.

  14. Solve each equation

    w=−132orw=5w = -\tfrac{13}{2} \quad \text{or} \quad w = 5

    Rearrange both linear equations.

  15. Reject the impossible value

    w=5w = 5

    A width cannot be negative, so w=5w = 5 cm (the length is then 1313 cm).

Answer
55
Question 4
6 markschallenging
Simplify fully 4x2−92x2−x−3\dfrac{4x^2 - 9}{2x^2 - x - 3}.
Show worked solution

Worked solution

  1. Plan the method

    numeratordenominator\frac{\text{numerator}}{\text{denominator}}

    Factorise the top and the bottom, then cancel any common bracket.

  2. Factorise the numerator

    4x2−0x−9=(2x+3)(2x−3)4x^2 - 0x - 9 = (2x + 3)(2x - 3)

    Use the AC method (or difference of two squares) on the numerator.

  3. Check the numerator

    (2x+3)(2x−3)=4x2−0x−9(2x + 3)(2x - 3) = 4x^2 - 0x - 9

    Expanding confirms the factors.

  4. Factorise the denominator

    2x2−1x−3=(2x−3)(x+1)2x^2 - 1x - 3 = (2x - 3)(x + 1)

    Factorise the bottom in the same way.

  5. Check the denominator

    (2x−3)(x+1)=2x2−1x−3(2x - 3)(x + 1) = 2x^2 - 1x - 3

    Expanding confirms these factors too.

  6. Rewrite the fraction

    (2x+3)(2x−3)(2x−3)(x+1)\frac{(2x + 3)(2x - 3)}{(2x - 3)(x + 1)}

    Replace the top and bottom with their factorised forms.

  7. Spot the common factor

    (2x−3)(2x - 3)

    The bracket (2x−3)(2x - 3) appears in both the numerator and the denominator.

  8. Cancel the common factor

    (2x+3)(x+1)\frac{(2x + 3)}{(x + 1)}

    Dividing top and bottom by (2x−3)(2x - 3) removes it.

  9. State the simplified expression

    (2x+3)(x+1)\frac{(2x + 3)}{(x + 1)}

    This is the fraction in its simplest form.

  10. Note the restriction

    x≠32, x≠−1x \neq \frac{3}{2}, \ x \neq -1

    The cancelled values are excluded because they make the original denominator zero.

  11. Why cancelling is allowed

    (2x−3)(2x−3)=1\frac{(2x - 3)}{(2x - 3)} = 1

    A bracket divided by itself equals one, provided it is not zero.

  12. Avoid the common mistake

    cancel factors, not individual terms\text{cancel factors, not individual terms}

    You may only cancel whole brackets, never single terms from a sum.

  13. Verify with a value

    x=1: 4(1)2−0(1)−92(1)2−1(1)−3x = 1: \ \frac{4(1)^2 - 0(1) - 9}{2(1)^2 - 1(1) - 3}

    Substitute x=1x = 1 into the original fraction.

  14. Confirm they match

    =52= \frac{5}{2}

    The simplified expression gives the same value at x=1x = 1.

  15. State the final answer

    (2x+3)(x+1)\frac{(2x + 3)}{(x + 1)}

    So the fraction simplifies to (2x+3)(x+1)\frac{(2x + 3)}{(x + 1)}.

Answer
(2x+3)(x+1)\frac{(2x + 3)}{(x + 1)}
Question 5
5 markschallenging
Factorise fully 2x3+7x2+3x2x^3 + 7x^2 + 3x.
Show worked solution

Worked solution

  1. Look for a common factor

    2x3+7x2+3x2x^3 + 7x^2 + 3x

    Every term contains at least one xx.

  2. Factor out xx

    x(2x2+7x+3)x(2x^2 + 7x + 3)

    Take the common factor xx outside the bracket.

  3. Set up the quadratic

    a=2,b=7,c=3a = 2, \quad b = 7, \quad c = 3

    Now factorise the quadratic in the bracket.

  4. Find the product acac

    2×3=62 \times 3 = 6

    Apply the AC method.

  5. List factor pairs

    6=1×6, 2×36 = 1\times6, \ 2\times3

    Consider the factor pairs of 66.

  6. Choose the pair

    6+1=7,6×1=66 + 1 = 7, \quad 6 \times 1 = 6

    The numbers 66 and 11 add to 77.

  7. Split the middle term

    2x2+6x+x+32x^2 + 6x + x + 3

    Rewrite 7x7x as 6x+x6x + x.

  8. Group into pairs

    (2x2+6x)+(x+3)(2x^2 + 6x) + (x + 3)

    Bracket the terms in twos.

  9. Factorise in pairs

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

    Both pairs leave (x+3)(x + 3).

  10. Factorise the quadratic

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

    Factor out the shared bracket.

  11. Combine with the xx

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

    Bring back the common factor xx.

  12. Check the quadratic

    (2x+1)(x+3)=2x2+7x+3(2x + 1)(x + 3) = 2x^2 + 7x + 3

    Expanding the two brackets is correct.

  13. Check the whole expression

    x(2x+1)(x+3)=2x3+7x2+3xx(2x + 1)(x + 3) = 2x^3 + 7x^2 + 3x

    Multiplying through by xx returns the original cubic.

  14. Confirm it is fully factorised

    three linear factors\text{three linear factors}

    Each factor is linear, so no further factorising is possible.

  15. State the answer

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

    This is the fully factorised form.

Answer
x(2x+1)(x+3)x(2x + 1)(x + 3)

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