GCSE Quadratics by factorising Practice Questions

Free GCSE Quadratics by factorising practice questions with full step-by-step worked solutions. Covers null-factor law, solving from brackets, common factor, zero root. Practise exam-style problems and check your method.

null-factor lawsolving from bracketscommon factorzero rootbracket with a coefficientfactorising a quadratic
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
Solve (x−3)(x+5)=0(x - 3)(x + 5) = 0.
Show worked solution

Worked solution

  1. Apply the null-factor law

    x−3=0orx+5=0x - 3 = 0 \quad \text{or} \quad x + 5 = 0

    The left-hand side is already a product of brackets equal to zero. A product is zero only if one of the parts is zero, so set each bracket equal to zero.

  2. Solve each bracket

    x=3orx=−5x = 3 \quad \text{or} \quad x = -5

    Each bracket is a simple linear equation. Notice how the signs flip: the numbers in the brackets are not the answers.

  3. State the solutions

    x=3 or x=−5x = 3 \text{ or } x = -5

    Two brackets give two solutions. You can check either one by putting it back into the original equation — it makes one bracket zero, so the product is zero.

Answer
x=3 or x=−5x = 3 \text{ or } x = -5
Question 2
1 markeasy
The equation (x−2)(x+8)=0(x - 2)(x + 8) = 0 has two solutions. Explain why.
Show worked solution

Worked solution

  1. Look at the structure of the equation

    (x−2)(x+8)=0(x - 2)(x + 8) = 0

    The left-hand side is two things multiplied together, and the result is zero.

  2. Apply the null-factor law

    x−2=0orx+8=0x - 2 = 0 \quad \text{or} \quad x + 8 = 0

    A product can only be zero if one of the factors is zero. That gives two separate linear equations — and it is "or", not "and": the brackets do not both have to be zero at once.

  3. Each bracket contributes one solution

    x=2orx=−8x = 2 \quad \text{or} \quad x = -8

    Two brackets, two ways of making the product zero, so two solutions.

Answer
x=2 or x=−8x = 2 \text{ or } x = -8
Question 3
2 marksintermediate
A student solves x2=5xx^2 = 5x by dividing both sides by xx to get x=5x = 5. Explain what is wrong with this method.
Show worked solution

Worked solution

  1. Never divide an equation by the unknown

    x2=5xx^2 = 5x

    Dividing both sides by x silently assumes x is not zero — but x=0x = 0 might be one of the answers, and here it is.

  2. Rearrange so that one side is zero

    x2−5x=0x^2 - 5x = 0

    Subtract 5x from both sides instead of dividing.

  3. Factorise

    x(x−5)=0x(x - 5) = 0

    Both terms contain x, so x comes out as a common factor.

  4. Apply the null-factor law

    x=0orx−5=0x = 0 \quad \text{or} \quad x - 5 = 0

    The factor x itself can be zero — that is the solution the division destroyed.

  5. Solve each bracket

    x=0orx=5x = 0 \quad \text{or} \quad x = 5

    Two factors, two solutions.

  6. Check the lost solution

    02=5×0⇒0=00^2 = 5 \times 0 \Rightarrow 0 = 0

    x=0x = 0 really does satisfy the original equation, so any method that misses it is incomplete. The student found only half the answer.

Answer
x=0 or x=5x = 0 \text{ or } x = 5
Question 4
4 markshard
A rectangle has length (x+3)(x + 3) cm and width xx cm. Its area is 5454 cm2^2. Find the value of xx.
Show worked solution

Worked solution

  1. Form an equation from the area

    x(x+3)=54x(x + 3) = 54

    Area of a rectangle = length × width, so x(x + 3) = 54.

  2. Expand and rearrange

    x2+3x=54⇒x2+3x−54=0x^2 + 3x = 54 \Rightarrow x^2 + 3x - 54 = 0

    Multiply out, then subtract 54 so that the equation equals zero.

  3. Look for the pair of numbers

    product=−54,sum=3\text{product} = -54, \quad \text{sum} = 3

    For x2x^2 + bx + c the two numbers in the brackets must multiply to c=−54c = -54 and add to b=3b = 3.

  4. Choose the pair that works

    (−6)×9=−54,(−6)+9=3(-6) \times 9 = -54, \quad (-6) + 9 = 3

    -6 and 9 pass both tests, so they are the numbers that go in the brackets.

  5. Write the factorised equation

    (x−6)(x+9)=0(x - 6)(x + 9) = 0

    The product of the brackets is zero — this is the form the null-factor law needs.

  6. Apply the null-factor law

    x−6=0orx+9=0x - 6 = 0 \quad \text{or} \quad x + 9 = 0

    If several things multiply to give zero, at least one of them must be zero. So set each bracket equal to zero in turn.

  7. Solve bracket 1

    x−6=0⇒x=6x - 6 = 0 \Rightarrow x = 6

    Solve the little linear equation x−6=0x - 6 = 0 to get x=6x = 6.

  8. Solve bracket 2

    x+9=0⇒x=−9x + 9 = 0 \Rightarrow x = -9

    Solve the little linear equation x+9=0x + 9 = 0 to get x=−9x = -9.

  9. State the solutions

    x=6 or x=−9x = 6 \text{ or } x = -9

    Each bracket gives one solution, so a quadratic that factorises into two different brackets has two solutions.

  10. Reject the impossible solution

    x=−9 is impossible, so x=6x = -9 \text{ is impossible, so } x = 6

    xx is a length, so it cannot be negative. Only x=6x = 6 makes sense: the rectangle is 99 cm by 66 cm, and 9×6=54 cm29 \times 6 = 54\text{ cm}^2 as required.

Answer
x=6x = 6
Question 5
6 markschallenging
Two positive numbers differ by 55. Their product is 8484. Find the two numbers.
Show worked solution

Worked solution

  1. Name the numbers

    n and n+5n \text{ and } n + 5

    The numbers differ by 5, so call the smaller one n and the larger one n + 5.

  2. Form the equation

    n(n+5)=84n(n + 5) = 84

    Their product is 84.

  3. Expand and rearrange

    n2+5n=84⇒n2+5n−84=0n^2 + 5n = 84 \Rightarrow n^2 + 5n - 84 = 0

    Multiply out, then subtract 84 so the equation equals zero.

  4. Look for the pair of numbers

    product=−84,sum=5\text{product} = -84, \quad \text{sum} = 5

    For n2n^2 + bn + c the two numbers in the brackets must multiply to c=−84c = -84 and add to b=5b = 5.

  5. List the factor pairs of -84

    (−84)×1,(−42)×2,(−28)×3,(−21)×4,(−14)×6,(−12)×7,(−7)×12,(−6)×14,(−4)×21,(−3)×28,(−2)×42,(−1)×84(-84) \times 1,\quad (-42) \times 2,\quad (-28) \times 3,\quad (-21) \times 4,\quad (-14) \times 6,\quad (-12) \times 7,\quad (-7) \times 12,\quad (-6) \times 14,\quad (-4) \times 21,\quad (-3) \times 28,\quad (-2) \times 42,\quad (-1) \times 84

    Write out every pair of whole numbers whose product is -84, then test which pair also adds to 5.

  6. Choose the pair that works

    (−7)×12=−84,(−7)+12=5(-7) \times 12 = -84, \quad (-7) + 12 = 5

    -7 and 12 pass both tests, so they are the numbers that go in the brackets.

  7. Write the factorised equation

    (n−7)(n+12)=0(n - 7)(n + 12) = 0

    The product of the brackets is zero — this is the form the null-factor law needs.

  8. Check the factorisation by expanding

    (n−7)(n+12)=n2+5n−84(n - 7)(n + 12) = n^2 + 5n - 84

    Multiplying the brackets back out returns the original quadratic, so the factorisation is right.

  9. Apply the null-factor law

    n−7=0orn+12=0n - 7 = 0 \quad \text{or} \quad n + 12 = 0

    If several things multiply to give zero, at least one of them must be zero. So set each bracket equal to zero in turn.

  10. Solve bracket 1

    n−7=0⇒n=7n - 7 = 0 \Rightarrow n = 7

    Solve the little linear equation n−7=0n - 7 = 0 to get n=7n = 7.

  11. Solve bracket 2

    n+12=0⇒n=−12n + 12 = 0 \Rightarrow n = -12

    Solve the little linear equation n+12=0n + 12 = 0 to get n=−12n = -12.

  12. State the solutions

    n=7 or n=−12n = 7 \text{ or } n = -12

    Each bracket gives one solution, so a quadratic that factorises into two different brackets has two solutions.

  13. Reject the impossible solution

    n=−12 is rejected, so n=7n = -12 \text{ is rejected, so } n = 7

    Both numbers must be positive, so the negative root cannot be the answer.

  14. Answer the question that was asked

    7 and 127 \text{ and } 12

    The numbers are n=7n = 7 and n+5=12n + 5 = 12. Check: 12−7=512 - 7 = 5 and 7×12=847 \times 12 = 84.

  15. Check the solution n=7n = 7

    (7)2+5(7)−84=49+35−84=0(7)^2 + 5(7) - 84 = 49 + 35 - 84 = 0

    Substituting n=7n = 7 back into the original quadratic gives 0, so this solution is correct.

Answer
7 and 127 \text{ and } 12

Unlock 65 more Quadratics by factorising questions

Create a free account to work through every GCSE Quadratics by factorising question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Quadratics by factorising practice

Hard QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsExpanding binomials