Hard GCSE Factorising quadratics Questions

Challenging, exam-style GCSE Factorising quadratics questions with worked solutions. Stretch yourself on the hardest solving quadratics, factorising, difference of two squares, coefficients problems.

solving quadraticsfactorisingdifference of two squarescoefficientscommon factorfactorising quadratics
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
Solve x2+6x+9=0x^2 + 6x + 9 = 0 by factorising.
Show worked solution

Worked solution

  1. Aim to factorise, then solve

    x2+6x+9=0x^2 + 6x + 9 = 0

    A product is zero only when one factor is zero, so factorise the left-hand side first.

  2. Confirm it is in standard form

    x2+6x+9=0x^2 + 6x + 9 = 0

    The equation already equals zero, ready to factorise.

  3. Identify the coefficients

    a=1,b=6,c=9a = 1,\quad b = 6,\quad c = 9

    Compare with x2x^2 + bx + c=0c = 0.

  4. State the method

    product=9,sum=6\text{product} = 9,\quad \text{sum} = 6

    Find two numbers that multiply to c and add to b.

  5. Decide the signs

    product >0\text{product } > 0

    Because the product is positive, both numbers are positive.

  6. List the factor pairs of 9

    1×9,3×31 \times 9,\quad 3 \times 3

    Write out the ways to make 9.

  7. Test the sums

    3+3=63 + 3 = 6

    The pair 3 and 3 gives the required sum.

  8. Confirm the product

    3×3=93 \times 3 = 9

    The same pair gives the required product.

  9. Write the factorised equation

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

    Rewrite the quadratic as a product of two brackets.

  10. Begin the expansion check

    x2+3x+3x+9x^2 + 3x + 3x + 9

    Multiply out to verify.

  11. Combine the middle terms

    x2+6x+9x^2 + 6x + 9

    This matches the original quadratic, so the factorisation is correct.

  12. Apply the zero-product property

    (x+3)=0or(x+3)=0(x + 3) = 0 \quad \text{or} \quad (x + 3) = 0

    At least one bracket must equal zero.

  13. Solve the first bracket

    x=−3x = -3

    Rearrange (x+3)=0(x + 3) = 0.

  14. Solve the second bracket

    x=−3x = -3

    Rearrange (x+3)=0(x + 3) = 0.

  15. State the repeated solution

    x=−3x = -3

    Both brackets are the same, so there is one repeated solution.

Answer
x=−3x = -3
Question 2
6 markschallenging
The product of two consecutive even numbers is 168168. Find the numbers.
Show worked solution

Worked solution

  1. Name the numbers

    n and n+2n \text{ and } n + 2

    Consecutive even numbers differ by 2.

  2. Write the product equation

    n(n+2)=168n(n + 2) = 168

    Their product is 168.

  3. Expand

    n2+2n=168n^2 + 2n = 168

    Multiply n through the bracket.

  4. Rearrange to standard form

    n2+2n−168=0n^2 + 2n - 168 = 0

    Subtract 168 from both sides.

  5. State the method

    product=−168,sum=2\text{product} = -168,\quad \text{sum} = 2

    Two numbers must multiply to -168 and add to 2.

  6. Decide the signs

    product <0\text{product } < 0

    One number is positive and one is negative.

  7. List useful factor pairs of 168

    2×84,6×28,12×142 \times 84,\quad 6 \times 28,\quad 12 \times 14

    Look for a pair close together, since the sum is small.

  8. Choose the pair

    14×(−12)=−168,14+(−12)=214 \times (-12) = -168,\quad 14 + (-12) = 2

    14 and -12 work.

  9. Factorise

    (n+14)(n−12)=0(n + 14)(n - 12) = 0

    Rewrite as a product of brackets.

  10. Check the expansion

    n2−12n+14n−168=n2+2n−168n^2 - 12n + 14n - 168 = n^2 + 2n - 168

    This matches the rearranged equation.

  11. Use the zero-product property

    (n+14)=0or(n−12)=0(n + 14) = 0 \quad \text{or} \quad (n - 12) = 0

    At least one bracket is zero.

  12. Solve for n

    n=−14orn=12n = -14 \quad \text{or} \quad n = 12

    These are the possible first numbers.

  13. Use n=12n = 12

    12 and 1412 \text{ and } 14

    The even numbers are 12 and 14.

  14. Use n=−14n = -14

    −14 and −12-14 \text{ and } -12

    The even numbers are -14 and -12.

  15. State the answer

    12 and 14or−14 and −1212 \text{ and } 14 \quad \text{or} \quad -14 \text{ and } -12

    Check: 12×14=16812 \times 14 = 168 and (−14)×(−12)=168(-14) \times (-12) = 168. ✓

Answer
12 and 14 (or -14 and -12)
Question 3
5 markschallenging
Solve x2−5x=14x^2 - 5x = 14.
Show worked solution

Worked solution

  1. Write the equation

    x2−5x=14x^2 - 5x = 14

    Start from the given equation.

  2. Rearrange to standard form

    x2−5x−14=0x^2 - 5x - 14 = 0

    Subtract 14 from both sides.

  3. Identify the coefficients

    a=1,b=−5,c=−14a = 1,\quad b = -5,\quad c = -14

    Compare with x2x^2 + bx + c=0c = 0.

  4. State the method

    product=−14,sum=−5\text{product} = -14,\quad \text{sum} = -5

    Two numbers must multiply to -14 and add to -5.

  5. Decide the signs

    product <0\text{product } < 0

    One number is positive and one is negative.

  6. List factor pairs of 14

    1×14,2×71 \times 14,\quad 2 \times 7

    Consider the ways to make 14.

  7. Choose the pair

    (−7)×2=−14,(−7)+2=−5(-7) \times 2 = -14,\quad (-7) + 2 = -5

    -7 and 2 work.

  8. Factorise

    (x−7)(x+2)=0(x - 7)(x + 2) = 0

    Rewrite as a product of brackets.

  9. Check the expansion

    x2+2x−7x−14=x2−5x−14x^2 + 2x - 7x - 14 = x^2 - 5x - 14

    This matches the rearranged equation.

  10. Use the zero-product property

    (x−7)=0or(x+2)=0(x - 7) = 0 \quad \text{or} \quad (x + 2) = 0

    At least one bracket is zero.

  11. Solve each bracket

    x=7orx=−2x = 7 \quad \text{or} \quad x = -2

    These are the two solutions.

  12. Check x=7x = 7

    49−35=1449 - 35 = 14

    The left-hand side equals 14. ✓

  13. Check x=−2x = -2

    4+10=144 + 10 = 14

    The left-hand side equals 14. ✓

  14. Note both are valid

    x=7 or x=−2x = 7 \text{ or } x = -2

    Both solutions satisfy the original equation.

  15. State the solutions

    x=7orx=−2x = 7 \quad \text{or} \quad x = -2

    These are the answers.

Answer
x=7 or x=−2x = 7 \text{ or } x = -2
Question 4
5 markschallenging
A square has area x2+12x+36x^2 + 12x + 36. Find an expression for the length of one side.
Show worked solution

Worked solution

  1. Link area to side length

    area=side2\text{area} = \text{side}^2

    For a square, the side is the square root of the area, so factorise to a perfect square.

  2. State the method

    product=36,sum=12\text{product} = 36,\quad \text{sum} = 12

    Two numbers must multiply to 36 and add to 12.

  3. Decide the signs

    product >0, sum >0\text{product } > 0,\ \text{sum } > 0

    Both numbers are positive.

  4. List factor pairs of 36

    1×36,2×18,3×12,4×9,6×61 \times 36,\quad 2 \times 18,\quad 3 \times 12,\quad 4 \times 9,\quad 6 \times 6

    Consider the ways to make 36.

  5. Choose the pair

    6×6=36,6+6=126 \times 6 = 36,\quad 6 + 6 = 12

    6 and 6 work.

  6. Write the brackets

    (x+6)(x+6)(x + 6)(x + 6)

    Both brackets are the same.

  7. Recognise the perfect square

    (x+6)(x+6)=(x+6)2(x + 6)(x + 6) = (x + 6)^2

    A repeated bracket is written as a square.

  8. Write the area as a square

    x2+12x+36=(x+6)2x^2 + 12x + 36 = (x + 6)^2

    The area is a perfect square.

  9. Picture the area model

    square split into x2, 6x, 6x, 36\text{square split into } x^2,\ 6x,\ 6x,\ 36

    A square of side (x + 6) splits into four regions that add to the area.

  10. Add the regions

    x2+6x+6x+36=x2+12x+36x^2 + 6x + 6x + 36 = x^2 + 12x + 36

    The four regions total the original area.

  11. Confirm the perfect square

    (x+6)2(x + 6)^2

    This matches the area, confirming the diagram.

  12. Take the square root

    side=(x+6)2\text{side} = \sqrt{(x + 6)^2}

    The side length is the square root of the area.

  13. Simplify the square root

    (x+6)2=x+6\sqrt{(x + 6)^2} = x + 6

    The square root of a square is the bracket itself.

  14. Check by squaring

    (x+6)2=x2+12x+36(x + 6)^2 = x^2 + 12x + 36

    Squaring the side returns the area. ✓

  15. State the side length

    x+6x + 6

    The length of one side is (x + 6).

Answer
(x+6)(x + 6)
Question 5
5 markschallenging
Factorise fully 3x2−483x^2 - 48.
Show worked solution

Worked solution

  1. Identify the terms

    3x2 and −483x^2 \text{ and } -48

    Look for a common factor first.

  2. Find the common factor

    HCF of 3 and 48=3\text{HCF of } 3 \text{ and } 48 = 3

    Both coefficients divide by 3.

  3. Factor out 3

    3(x2−16)3(x^2 - 16)

    Divide each term by 3.

  4. Check that step

    3⋅x2=3x2,3⋅(−16)=−483 \cdot x^2 = 3x^2,\quad 3 \cdot (-16) = -48

    Expanding confirms the factor of 3.

  5. Look at the bracket

    x2−16x^2 - 16

    It is a difference of two squares.

  6. Recognise the squares

    16=4216 = 4^2

    16 is 4 squared.

  7. Write x squared as a square

    x2=x2x^2 = x^2

    x2x^2 is already a perfect square.

  8. Identify a and b

    a=x,b=4a = x,\quad b = 4

    Use a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b).

  9. Factorise the bracket

    (x+4)(x−4)(x + 4)(x - 4)

    Apply the difference of two squares rule.

  10. Combine with the common factor

    3(x+4)(x−4)3(x + 4)(x - 4)

    Keep the 3 in front.

  11. Verify the bracket

    (x+4)(x−4)=x2−16(x + 4)(x - 4) = x^2 - 16

    The bracket expands correctly.

  12. Verify everything

    3(x2−16)=3x2−483(x^2 - 16) = 3x^2 - 48

    This returns the original expression.

  13. Emphasise fully factorised

    3(x+4)(x−4)3(x + 4)(x - 4)

    Common factor removed and the bracket factorised.

  14. Restate the method

    common factor, then difference of two squares\text{common factor, then difference of two squares}

    Always take out a common factor before looking for other patterns.

  15. State the answer

    3(x+4)(x−4)3(x + 4)(x - 4)

    This is the fully factorised form.

Answer
3(x+4)(x−4)3(x + 4)(x - 4)

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