What are sign slips in the brackets?

For x^2 - 2x - 15 the numbers must multiply to -15 and add to -2, giving (x - 5)(x + 3), not (x + 5)(x - 3).

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How do you expand brackets and factorise algebraic expressions, including common factors, the difference of two squares and trinomials?

Expanding brackets (including the product of two brackets) and factorising algebraic expressions using a common factor, the difference of two squares, and trinomials with unitary and non-unitary leading coefficients.

A focused answer to the SQA National 5 Mathematics algebra content, covering expanding single and double brackets, and the three factorising methods examined: common factor, difference of two squares, and factorising trinomials with both unitary and non-unitary leading coefficients.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Expanding brackets
Factorising: always check for a common factor first
Difference of two squares
Factorising trinomials
Choosing the right method
Examples in context
Try this

What this dot point is asking

The SQA wants you to expand brackets, including the product of two brackets, and factorise using the three methods examined at National 5: taking out a common factor, the difference of two squares, and factorising trinomials (quadratic expressions) with both unitary and non-unitary leading coefficients. Factorising is the reverse of expanding, and the command word "fully" means every possible factor must be removed.

Expanding brackets

To expand a single bracket, multiply each inside term by the term outside: $3(2x - 5) = 6x - 15$ . To expand the product of two brackets, multiply every term in the first by every term in the second, then collect like terms.

Expand and simplify

(x + 3)(x - 5)

Expanding two brackets means multiplying every term in the first bracket by every term in the second. With two terms in each bracket, this gives four products in total, which are then collected.

Step 1: Multiply every pair of terms

Take each term in $(x + 3)$ and multiply it by each term in $(x - 5)$ :

$x \times x = x^2$
$x \times (-5) = -5x$
$3 \times x = 3x$
$3 \times (-5) = -15$

Step 2: Collect like terms

The two $x$ terms can be combined: $-5x + 3x = -2x$ .

x^2 - 5x + 3x - 15 = x^2 - 2x - 15

Final answer: $(x + 3)(x - 5) = x^2 - 2x - 15$ .

Factorising: always check for a common factor first

The first step in any factorising question is to look for a factor common to every term and take it outside a bracket. This is the reverse of expanding a single bracket.

Factorise

6x^2 + 9x

Factorising by a common factor is the reverse of expanding a single bracket. We identify the largest factor that divides into every term, then write it outside a bracket.

Step 1: Find the highest common factor

Look for the largest number and the highest power of $x$ that divides both $6x^2$ and $9x$ . The highest common factor of $6$ and $9$ is $3$ ; both terms also contain at least one $x$ . So the highest common factor is $3x$ .

Step 2: Divide each term by the common factor and write the result

Dividing $6x^2$ by $3x$ gives $2x$ , and dividing $9x$ by $3x$ gives $3$ :

6x^2 + 9x = 3x(2x + 3)

Check by expanding: $3x \times 2x = 6x^2$ and $3x \times 3 = 9x$ . Correct.

Final answer: $6x^2 + 9x = 3x(2x + 3)$ .

Difference of two squares

When an expression is one square subtracted from another, with no middle term, it factorises into a sum and a difference.

For example, $x^2 - 49 = (x + 7)(x - 7)$ , and $9y^2 - 16 = (3y + 4)(3y - 4)$ because $9y^2 = (3y)^2$ and $16 = 4^2$ .

Factorising trinomials

A trinomial $x^2 + bx + c$ factorises into $(x + p)(x + q)$ where $p$ and $q$ multiply to $c$ and add to $b$ .

Factorise

x^2 + 7x + 12

A trinomial $x^2 + bx + c$ factorises into $(x + p)(x + q)$ where $p$ and $q$ multiply to $c$ and add to $b$ . We look for the pair of numbers that satisfies both conditions simultaneously.

Step 1: Identify the target product and sum

We need two numbers that multiply to $c = 12$ and add to $b = 7$ . List factor pairs of $12$ : $(1, 12)$ , $(2, 6)$ , $(3, 4)$ . Their sums are $13$ , $8$ and $7$ . The pair $(3, 4)$ works: $3 \times 4 = 12$ and $3 + 4 = 7$ .

Step 2: Write the factorised form

Replace $p = 3$ and $q = 4$ in $(x + p)(x + q)$ :

x^2 + 7x + 12 = (x + 3)(x + 4)

Check by expanding: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ . Correct.

Final answer: $x^2 + 7x + 12 = (x + 3)(x + 4)$ .

When the leading coefficient is not $1$ (a non-unitary trinomial like $2x^2 + 7x + 3$ ), multiply the first and last coefficients, find two numbers that multiply to that product and add to the middle coefficient, then split the middle term and factorise in pairs.

Factorise

2x^2 + 7x + 3

When the coefficient of $x^2$ is not $1$ , the simple "find two numbers adding to $b$ " shortcut no longer works directly. Instead, we multiply the first and last coefficients, find a pair of numbers for that product, then split the middle term and factorise in groups.

Step 1: Multiply the first and last coefficients

The first coefficient is $2$ and the last is $3$ , so the product is $2 \times 3 = 6$ . We need two numbers that multiply to $6$ and add to the middle coefficient $7$ . The pair $6$ and $1$ works: $6 \times 1 = 6$ and $6 + 1 = 7$ .

Step 2: Split the middle term using those two numbers

Rewrite $7x$ as $6x + x$ :

2x^2 + 6x + x + 3

Step 3: Factorise in pairs

Group the first two terms and the last two terms, taking out a common factor from each group:

2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

The bracket $(x + 3)$ is common to both groups, so it factors out.

Final answer: $2x^2 + 7x + 3 = (2x + 1)(x + 3)$ .

A useful check at every stage is to expand your answer back out. If $(2x + 1)(x + 3)$ does not give $2x^2 + 7x + 3$ when multiplied, you know a sign or a number is wrong. Because factorising and expanding are exact opposites, this self-check costs only a moment and catches most slips.

Choosing the right method

Many questions need more than one method, so it helps to have an order of attack. First, take out any common numerical or algebraic factor. Then look at what is left. If it has two terms and is a difference of squares, use $a^2 - b^2 = (a + b)(a - b)$ . If it has three terms (a trinomial), find the pair of numbers that multiply and add correctly, using the leading coefficient for non-unitary cases. A fully factorised answer cannot be broken down any further.

Factorise fully

3x^2 + 15x + 18

"Fully" means applying every possible method. Always remove a common factor first; this simplifies what is left and often reveals a straightforward trinomial.

Step 1: Take out the common numerical factor

Every coefficient ( $3$ , $15$ , $18$ ) is divisible by $3$ :

3x^2 + 15x + 18 = 3(x^2 + 5x + 6)

Step 2: Factorise the remaining trinomial

Now focus on $x^2 + 5x + 6$ . We need two numbers that multiply to $6$ and add to $5$ . The pair $2$ and $3$ works: $2 \times 3 = 6$ and $2 + 3 = 5$ . So:

x^2 + 5x + 6 = (x + 2)(x + 3)

Step 3: Write the complete factorisation

Combine the factor removed in Step 1 with the result from Step 2:

3x^2 + 15x + 18 = 3(x + 2)(x + 3)

Final answer: $3x^2 + 15x + 18 = 3(x + 2)(x + 3)$ .

Examples in context

Factorising is the gateway to solving quadratic equations and simplifying algebraic fractions. A rectangular garden whose area is $x^2 + 5x + 6$ square metres has dimensions $(x + 2)$ by $(x + 3)$ metres, found by factorising; this lets a designer read off the side lengths directly. Engineers factorise to locate where a quantity is zero, since a product is zero only when one of its factors is zero.

Try this

Q1. Expand and simplify $(2x - 1)(x + 4)$ . [2 marks]

Cue. $2x^2 + 7x - 4$ .

Q2. Factorise fully $x^2 - 25$ . [1 mark]

Cue. $(x + 5)(x - 5)$ .

Q3. Factorise $x^2 - x - 20$ . [2 marks]

Cue. $(x - 5)(x + 4)$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA National 5 20193 marksFactorise fully

2x^2 - 8

Show worked answer →

Take out the common factor first: $2x^2 - 8 = 2(x^2 - 4)$ (1 mark). Recognise $x^2 - 4$ as a difference of two squares, since $4 = 2^2$ (1 mark). Factorise it: $x^2 - 4 = (x + 2)(x - 2)$ , giving $2(x + 2)(x - 2)$ (1 mark). Markers reward removing the common factor and the full difference-of-squares factorisation. The word "fully" means the common factor must be taken out.

SQA National 5 20222 marksFactorise

3x^2 + 7x + 2

Show worked answer →

This is a non-unitary trinomial, so use the product $3 \times 2 = 6$ and find two numbers that multiply to $6$ and add to $7$ : these are $6$ and $1$ (1 mark). Split the middle term: $3x^2 + 6x + x + 2 = 3x(x + 2) + 1(x + 2) = (3x + 1)(x + 2)$ (1 mark). Markers reward a correct factorised pair of brackets; checking by expanding confirms the middle term is $7x$ .

Related dot points

Sources & how we know this

SQA National 5 Mathematics Course Specification — SQA (2018)