Quadratics by factorising Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Quadratics by factorising questions. See exactly how to solve problems on null-factor law, solving from brackets, common factor, zero root.

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.

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
Solve (x+2)(x+7)=0(x + 2)(x + 7) = 0.

Worked solution

  1. Apply the null-factor law

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

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

    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=−2 or x=−7x = -2 \text{ or } x = -7
Question 3
1 markeasy
Solve x(x−6)=0x(x - 6) = 0.

Worked solution

  1. Apply the null-factor law

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

    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=0 or x=6x = 0 \text{ or } x = 6

    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=0 or x=6x = 0 \text{ or } x = 6
Question 4
2 markseasy
Solve (2x−1)(x+3)=0(2x - 1)(x + 3) = 0.

Worked solution

  1. Apply the null-factor law

    2x−1=0orx+3=02x - 1 = 0 \quad \text{or} \quad x + 3 = 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=12orx=−3x = \frac{1}{2} \quad \text{or} \quad x = -3

    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=12 or x=−3x = \frac{1}{2} \text{ or } x = -3

    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=12 or x=−3x = \frac{1}{2} \text{ or } x = -3
Question 5
2 markseasy
Solve x2+5x+4=0x^2 + 5x + 4 = 0 by factorising.

Worked solution

  1. Factorise the left-hand side

    x2+5x+4=(x+1)(x+4)x^2 + 5x + 4 = (x + 1)(x + 4)

    Two numbers multiply to 4 and add to 5: 1 and 4. They go straight into the brackets.

  2. Apply the null-factor law

    x+1=0orx+4=0x + 1 = 0 \quad \text{or} \quad x + 4 = 0

    The brackets multiply to give zero, so at least one of them must be zero. Set each one equal to zero.

  3. Solve each bracket and state the solutions

    x=−1 or x=−4x = -1 \text{ or } x = -4

    Solving the two little linear equations gives the two solutions. Check one by substituting it back into the original equation.

Answer
x=−1 or x=−4x = -1 \text{ or } x = -4

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