GCSE Factorising quadratics Practice Questions

Free GCSE Factorising quadratics practice questions with full step-by-step worked solutions. Covers common factor, factorising, factorising quadratics, factor pairs. Practise exam-style problems and check your method.

common factorfactorisingfactorising quadraticsfactor pairsdifference of two squaresperfect square
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Factorise 2x+62x + 6.
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Worked solution

  1. Find the highest common factor

    HCF=2\text{HCF} = 2

    Every term divides by 22.

  2. Divide each term by 22

    2x÷2=x,6÷2=32x \div 2 = x,\quad 6 \div 2 = 3

    Write what is left inside the bracket.

  3. Write the factorised form

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

    Check: 2×x=2x2 \times x = 2x and 2×3=62 \times 3 = 6, giving 2x+62x + 6.

Answer
2(x+3)2(x + 3)
Question 2
1 markeasy
Factorise 6x+96x + 9.
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Worked solution

  1. Find the highest common factor

    HCF=3\text{HCF} = 3

    Every term divides by 33.

  2. Divide each term by 33

    6x÷3=2x,9÷3=36x \div 3 = 2x,\quad 9 \div 3 = 3

    Write what is left inside the bracket.

  3. Write the factorised form

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

    Check: 3×2x=6x3 \times 2x = 6x and 3×3=93 \times 3 = 9.

Answer
3(2x+3)3(2x + 3)
Question 3
2 marksintermediate
Factorise fully 2x2−2x−122x^2 - 2x - 12.
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Worked solution

  1. Look for a common factor

    HCF=2\text{HCF} = 2

    Every term divides by 22, so take it out first.

  2. Factor out 22

    2(x2−x−6)2(x^2 - x - 6)

    Divide each term by 22.

  3. Factorise the bracket

    product=−6,sum=−1\text{product} = -6,\quad \text{sum} = -1

    Two numbers must multiply to −6-6 and add to −1-1.

  4. Choose the pair

    (−3)×2=−6,(−3)+2=−1(-3) \times 2 = -6,\quad (-3) + 2 = -1

    −3-3 and 22 work.

  5. Write the full factorisation

    2(x−3)(x+2)2(x - 3)(x + 2)

    Keep the common factor of 22 in front.

  6. Check by expanding

    2(x2−x−6)=2x2−2x−122(x^2 - x - 6) = 2x^2 - 2x - 12

    Expanding returns the original expression.

Answer
2(x−3)(x+2)2(x - 3)(x + 2)
Question 4
3 markshard
Solve x2−3x−10=0x^2 - 3x - 10 = 0 by factorising.
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Worked solution

  1. Aim to factorise, then solve

    x2−3x−10=0x^2 - 3x - 10 = 0

    A product equals zero when one of its factors is zero, so factorise first.

  2. Identify the coefficients

    b=−3,c=−10b = -3,\quad c = -10

    Compare with x2x^2 + bx + c.

  3. State what is needed

    product=−10,sum=−3\text{product} = -10,\quad \text{sum} = -3

    Two numbers multiply to −10-10 and add to −3-3.

  4. Decide the signs

    product <0\text{product } < 0

    Because the product is negative, one number is positive and one is negative.

  5. Find the pair

    (−5)×2=−10,(−5)+2=−3(-5) \times 2 = -10,\quad (-5) + 2 = -3

    −5-5 and 22 work.

  6. Factorise

    (x−5)(x+2)=0(x - 5)(x + 2) = 0

    Rewrite the equation in factorised form.

  7. Check the expansion

    x2+2x−5x−10=x2−3x−10x^2 + 2x - 5x - 10 = x^2 - 3x - 10

    This matches the original quadratic.

  8. Use the zero-product property

    (x−5)=0or(x+2)=0(x - 5) = 0 \quad \text{or} \quad (x + 2) = 0

    If two things multiply to zero, at least one must be zero.

  9. Solve the first bracket

    x=5x = 5

    Rearrange (x−5)=0(x - 5) = 0.

  10. Solve the second bracket

    x=−2x = -2

    Rearrange (x+2)=0(x + 2) = 0.

Answer
x=5 or x=−2x = 5 \text{ or } x = -2
Question 5
5 markschallenging
Solve x2+6x+9=0x^2 + 6x + 9 = 0 by factorising.
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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 99

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

    Write out the ways to make 99.

  7. Test the sums

    3+3=63 + 3 = 6

    The pair 33 and 33 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

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