Year 10 - Quadratic factorising - Free practice

Year 10 Quadratic Factorising Worksheets

Expand, factorise, difference of squares - the building blocks.

Quick answer

How do you factorise quadratic expressions?

To factorise a monic quadratic like x² + bx + c, find two numbers that multiply to c and add to b, then write it as (x + p)(x + q). For a difference of two squares, use a² - b² = (a + b)(a - b). For example, x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 10 quadratic factorising: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

  1. Example 1 · Factorising x² + bx + c

    Factorise x2+7x+12x^2 + 7x + 12.

    Show worked solution

    Key idea: Find two integers that multiply to 1212 and add to 77.

    1. List factor pairs of 1212: (1,12), (2,6), (3,4)(1, 12),\ (2, 6),\ (3, 4).
    2. Check which pair adds to 77: 3+4=73 + 4 = 7. ✓
    3. Write the factorised form: (x+3)(x+4)(x + 3)(x + 4).
    4. Verify by expanding: (x+3)(x+4)=x2+4x+3x+12=x2+7x+12(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12. ✓

    Answer: (x+3)(x+4)(x + 3)(x + 4)

  2. Example 2 · Difference of two squares and perfect-square trinomials

    Factorise each expression.

    (a) x2−49x^2 - 49

    (b) x2−10x+25x^2 - 10x + 25

    Show worked solution

    Key idea: Recognise special quadratic forms: a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b) and a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2.

    Part (a)

    1. Write x2−49x^2 - 49 as x2−72x^2 - 7^2.
    2. Apply the difference of two squares rule: (x−7)(x+7)(x - 7)(x + 7).

    Part (b)

    1. Check: 25=5\sqrt{25} = 5 and the middle term −10x=−2×x×5-10x = -2 \times x \times 5. ✓
    2. This is a perfect-square trinomial: (x−5)2(x - 5)^2.
    3. Verify: (x−5)2=x2−10x+25(x-5)^2 = x^2 - 10x + 25. ✓

    Answer: (a) (x−7)(x+7)(x - 7)(x + 7) (b) (x−5)2(x - 5)^2

  3. Example 3 · Real-life area problem

    A rectangular garden has an area of x2+11x+28x^2 + 11x + 28 square metres, where xx is a positive number. The length of the garden is (x+7)(x + 7) metres. Write an expression for the width of the garden by factorising the area expression, then find the width when x=3x = 3.

    Show worked solution

    Key idea: Area of a rectangle equals length multiplied by width, so width equals area divided by length. Factorise the area expression first.

    1. Find two integers that multiply to 2828 and add to 1111: 4+7=114 + 7 = 11 and 4×7=284 \times 7 = 28. ✓
    2. Factorise: x2+11x+28=(x+4)(x+7)x^2 + 11x + 28 = (x + 4)(x + 7).
    3. Width =(x+4)(x+7)(x+7)=x+4= \dfrac{(x + 4)(x + 7)}{(x + 7)} = x + 4 metres.
    4. Substitute x=3x = 3: width =3+4=7= 3 + 4 = 7 metres.

    Answer: Width =(x+4)= (x + 4) metres; when x=3x = 3, the width is 77 metres.

Common mistakes

Where students usually lose marks

Wrong signs when factorising

x2−5x+6=(x+2)(x+3)x^2 - 5x + 6 = (x + 2)(x + 3) (ignoring the negative signs)

Always check both the product and the sum with their signs. Here you need two numbers that multiply to +6+6 and add to −5-5, which are −2-2 and −3-3, giving (x−2)(x−3)(x - 2)(x - 3). Expand your answer to check.

Difference of two squares applied to a sum

x2+49=(x−7)(x+7)x^2 + 49 = (x - 7)(x + 7) (treating a sum as a difference of squares)

The difference of two squares rule only works when there is a subtraction: a2−b2a^2 - b^2. The expression x2+49x^2 + 49 has no real factors and cannot be factorised over the integers.

Forgetting to check the middle term for a perfect square

x2+12x+36x^2 + 12x + 36 written as (x+6)(x+7)(x + 6)(x + 7) because 6×7=426 \times 7 = 42, not 3636

Before using the standard method, check whether the expression is a perfect-square trinomial: is the last term a perfect square, and does the middle term equal 2×x×last term2 \times x \times \sqrt{\text{last term}}? Here 2×6=122 \times 6 = 12 ✓, so the answer is (x+6)2(x + 6)^2.

Not fully factorising

2x2+6x+42x^2 + 6x + 4 factorised as (x+1)(2x+4)(x + 1)(2x + 4), leaving a common factor inside

Always take out any common numerical factor first. Here, take out 22: 2(x2+3x+2)=2(x+1)(x+2)2(x^2 + 3x + 2) = 2(x + 1)(x + 2). Check every bracket to make sure no further factorisation is possible.

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Your turn

Factorise x2−3x−28x^2 - 3x - 28 and find the positive value of xx for which the expression equals zero.

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About this worksheet

Why we built it

Year 10 expects students to fluently expand and factorise quadratics. These worksheets give the structured repetition needed to make the pattern recognition automatic - critical for senior Methods maths.

What's covered

Sub-skills your student will practise

  • ✓Expanding binomial products (FOIL)
  • ✓Factorising x^2 + bx + c
  • ✓Difference of two squares
  • ✓Perfect-square trinomials
  • ✓Mixed-format expand and factorise problems

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Frequently asked

Questions parents ask about quadratic factorising

What's the cleanest way to factorise a monic quadratic?

Find two numbers that multiply to the constant term and add to the x coefficient. For x^2 - 7x + 12: factors of 12 that add to -7 are -3 and -4. So (x - 3)(x - 4).

How do students factorise a non-monic quadratic?

Use the 'ac method' (or 'splitting the middle term'): for ax^2 + bx + c, find factors of ac that sum to b, then split bx into those two terms and factorise by grouping. Year 10A goes deeper into this.

What's perfect-square factorisation?

A specific pattern: a^2 + 2ab + b^2 = (a + b)^2 and a^2 - 2ab + b^2 = (a - b)^2. So x^2 + 6x + 9 factorises as (x + 3)^2. Recognising the pattern saves time on exam questions.

Why is fluent factorising so important in Year 10?

Because every Methods topic from now on assumes it - solving equations, sketching parabolas, finding intercepts, completing the square. Students who can't factorise fluently spend twice as long on every Methods question.

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