Expanding and Factorising Quadratics
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Revision notes for CIE IGCSE Maths Expanding and Factorising Quadratics. Open the guide for explanations and worked examples. Written against the CIE IGCSE Maths (0580) specification, so the content matches what's examinable rather than general Maths background.

Expanding and Factorising Quadratics

What you'll learn

  • How to expand two brackets and simplify the answer.
  • How signs affect quadratic expressions.
  • How to factorise quadratics of the form x2+bx+cx^2 + bx + cx2+bx+c.
  • How to spot special cases like squares and difference of two squares.

The key language

Before we start, make sure these words feel familiar.

Definition

Term, coefficient, and quadratic

  • A term is one part of an expression, such as x2x^2x2, 5x5x5x, or −12-12−12.
  • A coefficient is the number multiplying a variable. In 7x7x7x, the coefficient is 7.
  • A quadratic expression has a highest power of 2, such as x2+6x+8x^2 + 6x + 8x2+6x+8.

Quadratics often appear when you multiply two brackets together. For example, multiplying (x+4)(x+3)(x + 4)(x + 3)(x+4)(x+3) creates an x2x^2x2 term, so the answer is quadratic.

Prerequisite: multiplying algebra terms

When expanding brackets, you need to multiply numbers, variables, and signs accurately.

Useful rules:

  • x⋅x=x2x \cdot x = x^2x⋅x=x2
  • positive times positive gives positive
  • negative times negative gives positive
  • positive times negative gives negative
Example

Multiplying algebra terms

Simplify −3x⋅5x-3x \cdot 5x−3x⋅5x.

  1. Multiply the numbers first:

    −3⋅5=−15-3 \cdot 5 = -15−3⋅5=−15

  2. Multiply the variables:

    x⋅x=x2x \cdot x = x^2x⋅x=x2

  3. Put the parts together:

    −3x⋅5x=−15x2-3x \cdot 5x = -15x^2−3x⋅5x=−15x2

Common Mistake

Forgetting x⋅x=x2

A very common error is writing x⋅x=2xx \cdot x = 2xx⋅x=2x. That is not correct. Multiplying xxx by xxx gives x2x^2x2.

Expanding two brackets

Definition

Expand

To expand means to remove brackets by multiplying out. For two brackets, every term in the first bracket must multiply every term in the second bracket.

The safest method is the grid method or the “four products” method. Both do the same thing: they make sure nothing is missed.

This area model shows why four products appear when you expand two brackets.

Area model showing how (x + 4)(x + 3) expands into x squared, 4x, 3x and 12

Key Idea

Every term multiplies every term

For (x+a)(x+b)(x + a)(x + b)(x+a)(x+b), you get four products first, then you collect like terms.

Example: expanding simple brackets

Example

Expand and simplify (x+5)(x−2)

  1. Multiply the first terms:

    x⋅x=x2x \cdot x = x^2x⋅x=x2

  2. Multiply the outside terms:

    x⋅(−2)=−2xx \cdot (-2) = -2xx⋅(−2)=−2x

  3. Multiply the inside terms:

    5⋅x=5x5 \cdot x = 5x5⋅x=5x

  4. Multiply the last terms:

    5⋅(−2)=−105 \cdot (-2) = -105⋅(−2)=−10

  5. Write all four terms:

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

  6. Collect the like terms −2x-2x−2x and 5x5x5x:

    x2+3x−10x^2 + 3x - 10x2+3x−10

Tip

Quick check

After expanding two linear brackets, you should usually have an x2x^2x2 term, an xxx term, and a number term. If one is missing, check whether it has cancelled out rather than assuming it disappeared by accident.

Expanding when there are coefficients

A coefficient in front of a variable, such as the 3 in 3x3x3x, changes the first product and the middle terms. These questions are still just four multiplications.

Example: expanding with coefficients

Example

Expand and simplify (3p−4)(p−6)

  1. Multiply the first terms:

    3p⋅p=3p23p \cdot p = 3p^23p⋅p=3p2

  2. Multiply the outside terms:

    3p⋅(−6)=−18p3p \cdot (-6) = -18p3p⋅(−6)=−18p

  3. Multiply the inside terms:

    −4⋅p=−4p-4 \cdot p = -4p−4⋅p=−4p

  4. Multiply the last terms:

    −4⋅(−6)=24-4 \cdot (-6) = 24−4⋅(−6)=24

  5. Write the four products:

    3p2−18p−4p+243p^2 - 18p - 4p + 243p2−18p−4p+24

  6. Collect the like terms:

    3p2−22p+243p^2 - 22p + 243p2−22p+24

Common Mistake

Dropping the negative sign

In (3p−4)(p−6)(3p - 4)(p - 6)(3p−4)(p−6), the last product is (−4)(−6)(-4)(-6)(−4)(−6), which is positive 24. Treat subtraction signs as part of the term that follows them.

Expanding a squared bracket

A squared bracket means the bracket is multiplied by itself.

Definition

Squared bracket

(a−5)2(a - 5)^2(a−5)2 means (a−5)(a−5)(a - 5)(a - 5)(a−5)(a−5), not a2−25a^2 - 25a2−25.

Example: expanding a squared bracket

Example

Expand and simplify (y−8)2

  1. Rewrite the square as two identical brackets:

    (y−8)2=(y−8)(y−8)(y - 8)^2 = (y - 8)(y - 8)(y−8)2=(y−8)(y−8)

  2. Multiply the first terms:

    y⋅y=y2y \cdot y = y^2y⋅y=y2

  3. Multiply the outside and inside terms:

    y⋅(−8)=−8yy \cdot (-8) = -8yy⋅(−8)=−8y

    −8⋅y=−8y-8 \cdot y = -8y−8⋅y=−8y

  4. Multiply the last terms:

    (−8)(−8)=64(-8)(-8) = 64(−8)(−8)=64

  5. Collect the like terms:

    y2−8y−8y+64=y2−16y+64y^2 - 8y - 8y + 64 = y^2 - 16y + 64y2−8y−8y+64=y2−16y+64

Common Mistake

The middle term matters

(y−8)2(y - 8)^2(y−8)2 is y2−16y+64y^2 - 16y + 64y2−16y+64, not y2+64y^2 + 64y2+64. The middle term only disappears in cases like (y+8)(y−8)(y + 8)(y - 8)(y+8)(y−8).

Factorising quadratics

Definition

Factorise

To factorise means to put an expression back into brackets. It is the reverse of expanding.

For Grade 5 questions, you often factorise quadratics where the coefficient of x2x^2x2 is 1, such as:

x2+9x+20x^2 + 9x + 20x2+9x+20

You are looking for two numbers that:

  • multiply to make the final number
  • add to make the coefficient of xxx

For x2+bx+cx^2 + bx + cx2+bx+c, you need two numbers with product ccc and sum bbb.

Example: factorising when both signs are positive

Example

Factorise x2+11x+24

  1. Look for two numbers that multiply to 24.

    Possible pairs include 1 and 24, 2 and 12, 3 and 8, and 4 and 6.

  2. Choose the pair that adds to 11.

    3+8=113 + 8 = 113+8=11

  3. Write the brackets using those two numbers:

    (x+3)(x+8)(x + 3)(x + 8)(x+3)(x+8)

  4. Check by expanding mentally: the middle terms would be 3x+8x=11x3x + 8x = 11x3x+8x=11x.

Factorising with negative signs

The signs give you clues:

  • If the final number is positive and the middle term is positive, both signs are positive.
  • If the final number is positive and the middle term is negative, both signs are negative.
  • If the final number is negative, one sign is positive and one sign is negative.

Example: factorising with two negatives

Example

Factorise x2−9x+20

  1. The final number is positive 20, so the two numbers have the same sign.

  2. The middle term is negative, so both numbers must be negative.

  3. Find two numbers that multiply to 20 and add to 9.

    4+5=94 + 5 = 94+5=9

  4. Use negative 4 and negative 5:

    (x−4)(x−5)(x - 4)(x - 5)(x−4)(x−5)

  5. Check the middle term:

    −4x−5x=−9x-4x - 5x = -9x−4x−5x=−9x

Example: factorising with one positive and one negative

Example

Factorise x2+2x−35

  1. The final number is negative, so the bracket signs must be different.

  2. Find a factor pair of 35 with a difference of 2.

    5 and 7 have a difference of 2.

  3. Since the middle term is positive, the larger number must be positive.

  4. Write the brackets:

    (x+7)(x−5)(x + 7)(x - 5)(x+7)(x−5)

  5. Check the middle term:

    7x−5x=2x7x - 5x = 2x7x−5x=2x

Tip

Use product and sum

For x2+bx+cx^2 + bx + cx2+bx+c, say to yourself: “multiply to the end, add to the middle.” This is the fastest way to choose the bracket numbers.

Special case: difference of two squares

Some expressions factorise very quickly.

Definition

Difference of two squares

A difference of two squares has the form x2−a2x^2 - a^2x2−a2. It factorises as (x+a)(x−a)(x + a)(x - a)(x+a)(x−a).

This works because the middle terms cancel:

(x+a)(x−a)=x2−a2(x + a)(x - a) = x^2 - a^2(x+a)(x−a)=x2−a2

Example: factorising a difference of two squares

Example

Factorise x2−64

  1. Recognise both terms as squares:

    x2x^2x2 is a square, and 64 is 828^282.

  2. Write one bracket with plus and one bracket with minus:

    (x+8)(x−8)(x + 8)(x - 8)(x+8)(x−8)

  3. Check that the middle terms cancel:

    −8x+8x=0-8x + 8x = 0−8x+8x=0

Common Mistake

Only works with subtraction

x2−64x^2 - 64x2−64 factorises as (x+8)(x−8)(x + 8)(x - 8)(x+8)(x−8). But x2+64x^2 + 64x2+64 does not factorise like this at IGCSE level.

Special case: perfect square trinomials

A trinomial is an expression with three terms. Some quadratics come from squaring a bracket:

(x+a)2=x2+2ax+a2(x + a)^2 = x^2 + 2ax + a^2(x+a)2=x2+2ax+a2

and

(x−a)2=x2−2ax+a2(x - a)^2 = x^2 - 2ax + a^2(x−a)2=x2−2ax+a2

Example: factorising a perfect square

Example

Factorise x2+14x+49

  1. Notice that 49 is a square:

    49=7249 = 7^249=72

  2. Check the middle term: twice 7 is 14.

  3. Since the middle term is positive, use plus signs:

    (x+7)2(x + 7)^2(x+7)2

  4. You can also write this as:

    (x+7)(x+7)(x + 7)(x + 7)(x+7)(x+7)

Checking your factorisation

A quick check is to expand your answer. If it returns to the original expression, your factorisation is correct.

Example

Check whether (n+6)(n−4) matches n2+2n−24

  1. Expand the brackets:

    n2−4n+6n−24n^2 - 4n + 6n - 24n2−4n+6n−24

  2. Collect like terms:

    n2+2n−24n^2 + 2n - 24n2+2n−24

  3. Compare with the original expression.

    The expressions match, so the factorisation is correct.

Exam technique

In the exam

  1. For expanding, write all four products before collecting like terms.

  2. For factorising x2+bx+cx^2 + bx + cx2+bx+c, find two numbers that multiply to ccc and add to bbb.

  3. Always check signs carefully, especially when the final number is negative or when both brackets contain subtraction.

Self review

Check yourself

  • Can you expand (x−4)(x+9)(x - 4)(x + 9)(x−4)(x+9) without missing any of the four products?
  • Can you factorise x2−6x+8x^2 - 6x + 8x2−6x+8 by finding the correct product and sum?
  • Can you spot why x2−121x^2 - 121x2−121 is a difference of two squares?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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