Topics

Expanding and factorising

Covers AS91261

Expanding brackets

Expanding means multiplying out brackets so no brackets remain.
Every term inside the bracket is multiplied by what is outside:

$a(b + c) = ab + ac$

$3x(2x - 5) = 6x^2 - 15x$
Watch the signs. $-2(x - 4) = -2x + 8$ — the minus multiplies both terms.

Two brackets

Multiply every term in the first bracket by every term in the second. Four products for two binomials.

$(a+b)(c+d) = ac + ad + bc + bd$

$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$
The two patterns worth recognising instantly:

$(a + b)^2 = a^2 + 2ab + b^2$

$(a - b)^2 = a^2 - 2ab + b^2$

$(a+b)(a-b) = a^2 - b^2$

$(a+b)^2$ is not $a^2 + b^2$ . There is a middle term, and forgetting it is the most common error in the whole standard.
$(2y)^3 = 8y^3$ , not $2y^3$ — the power applies to the coefficient too. The 2025 report lists "expanded $(2y)^3$ to give $2y^3$ rather than $8y^3$ " as the first Not Achieved behaviour on this standard.

Factorising — common factors

Factorising is expanding in reverse: writing an expression as a product.
Always look for a common factor first.
- Find the highest number that divides every term.
- Find the lowest power of each letter appearing in every term.
- Take both outside the bracket.
$12x^3 - 18x^2 = 6x^2(2x - 3)$
Check by expanding back. If you do not recover the original, the factorisation is wrong.

Factorising quadratics

For $x^2 + bx + c$ (with a leading coefficient of 1):
- Find two numbers that multiply to $c$ and add to $b$ .
- Write them in the brackets.
$x^2 + 7x + 12$ : two numbers multiplying to 12 and adding to 7 are 3 and 4, so $(x+3)(x+4)$ .
$x^2 - 2x - 15$ : multiplying to $-15$ , adding to $-2$ , gives $-5$ and $3$ , so $(x-5)(x+3)$ .
The sign rules that speed this up:
- $c$ positive — both numbers have the same sign, matching $b$
- $c$ negative — the numbers have opposite signs

When the leading coefficient is not 1

For $ax^2 + bx + c$ :
- Multiply $a$ by $c$ .
- Find two numbers multiplying to $ac$ and adding to $b$ .
- Split the middle term using them.
- Factorise in pairs.
Worked through for $6x^2 + 11x + 3$ :
- $ac = 18$ ; two numbers multiplying to 18 and adding to 11 are 9 and 2
- Split: $6x^2 + 9x + 2x + 3$
- Pair: $3x(2x + 3) + 1(2x + 3)$
- Factor out the common bracket: $(2x+3)(3x+1)$

The difference of two squares

$a^2 - b^2 = (a+b)(a-b)$

Only works for a subtraction. $a^2 + b^2$ does not factorise over the real numbers.
$9x^2 - 25 = (3x+5)(3x-5)$
$x^4 - 16 = \left(x^2+4\right)\left(x^2-4\right) = \left(x^2+4\right)(x+2)(x-2)$ — factorise fully, applying the rule again where possible.

Completing the square

Rewriting $x^2 + bx + c$ $x^{2} + b x + c$ as $(x + p)^2 + q$ $(x + p)^{2} + q$ :
- Halve the coefficient of $x$ to get $p$ .
- Square it and subtract it, then add the original constant.

$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c$

$x^2 + 6x + 11 = (x+3)^2 - 9 + 11 = (x+3)^2 + 2$
This form shows the vertex immediately: the minimum of $(x+3)^2 + 2$ is at $(-3, 2)$ , because a square is never negative.

Tips and Tricks

The power applies to everything inside the bracket. $(2y)^3 = 2^3y^3 = 8y^3$ and $(3x)^2 = 9x^2$ . The assessor report names this exact error first among the reasons candidates fail this standard.

$(a+b)^2$ has three terms. Write out the four products the first few times rather than jumping to a memorised shortcut — the middle term is where the marks go.

Check every factorisation by expanding it back. It takes one line and it is the only way to be certain. The 2025 report lists "factorised quadratics" as an Achieved behaviour and "lacked basic algebraic skills (factorising, substitution etc)" among the Not Achieved.

Take out the common factor first, always. $2x^2 - 8 = 2(x^2 - 4) = 2(x+2)(x-2)$ is easy; trying to factorise $2x^2 - 8$ directly is not.

Factorise fully. Stopping at $\left(x^2+4\right)\left(x^2-4\right)$ when the second bracket still factorises loses the mark.

The report also lists "incorrectly dealt with negative numbers" and "substituted incorrectly into algebraic expressions". Use brackets when substituting: to find the value of $-x^2$ at $x = -3$ , write $-(-3)^2 = -9$ , not $9$ .

Worked ExampleFactorising fully

Factorise completely: (a) $3x^2 - 27$ , (b) $2x^2 - 7x - 15$ .

Part (a), Step 1 — Look for a common factor first

Both terms are divisible by 3:

$3x^2 - 27 = 3\left(x^2 - 9\right)$

Part (a), Step 2 — Recognise the difference of two squares

Inside the bracket, $x^2 - 9$ is $x^2 - 3^2$ , a difference of two squares:

$a^2 - b^2 = (a+b)(a-b) \quad \text{with } a = x, \; b = 3$

$= 3(x+3)(x-3)$

Part (a), Step 3 — Check by expanding

$3(x+3)(x-3) = 3\left(x^2 - 9\right) = 3x^2 - 27 \;\checkmark$

Part (b), Step 1 — Check for a common factor

$2$ , $7$ and $15$ share no common factor, so there is nothing to take out. The leading coefficient is not 1, so use the splitting method.

Part (b), Step 2 — Multiply $a$ by $c$

$a \times c = 2 \times (-15) = -30$

Part (b), Step 3 — Find two numbers

We need two numbers that multiply to $-30$ and add to $-7$ (the coefficient of $x$ ).

Since the product is negative, the two numbers have opposite signs. Testing pairs:

$-10$ and $3$ : product $-30$ ✓, sum $-7$ ✓

Part (b), Step 4 — Split the middle term

Replace $-7x$ with $-10x + 3x$ :

$2x^2 - 10x + 3x - 15$

Part (b), Step 5 — Factorise in pairs

Group the first two and the last two, and take a common factor from each:

$2x(x - 5) + 3(x - 5)$

Part (b), Step 6 — Take out the common bracket

$(x - 5)(2x + 3)$

Part (b), Step 7 — Check by expanding

$(x-5)(2x+3) = 2x^2 + 3x - 10x - 15 = 2x^2 - 7x - 15 \;\checkmark$

Expanding and factorising

Covers AS91261

Expanding brackets

Expanding means multiplying out brackets so no brackets remain.
Every term inside the bracket is multiplied by what is outside:

$a(b + c) = ab + ac$

$3x(2x - 5) = 6x^2 - 15x$
Watch the signs. $-2(x - 4) = -2x + 8$ — the minus multiplies both terms.

Two brackets

Multiply every term in the first bracket by every term in the second. Four products for two binomials.

$(a+b)(c+d) = ac + ad + bc + bd$

$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$
The two patterns worth recognising instantly:

$(a + b)^2 = a^2 + 2ab + b^2$

$(a - b)^2 = a^2 - 2ab + b^2$

$(a+b)(a-b) = a^2 - b^2$

$(a+b)^2$ is not $a^2 + b^2$ . There is a middle term, and forgetting it is the most common error in the whole standard.
$(2y)^3 = 8y^3$ , not $2y^3$ — the power applies to the coefficient too. The 2025 report lists "expanded $(2y)^3$ to give $2y^3$ rather than $8y^3$ " as the first Not Achieved behaviour on this standard.

Factorising — common factors

Factorising is expanding in reverse: writing an expression as a product.
Always look for a common factor first.
- Find the highest number that divides every term.
- Find the lowest power of each letter appearing in every term.
- Take both outside the bracket.
$12x^3 - 18x^2 = 6x^2(2x - 3)$
Check by expanding back. If you do not recover the original, the factorisation is wrong.

Factorising quadratics

For $x^2 + bx + c$ (with a leading coefficient of 1):
- Find two numbers that multiply to $c$ and add to $b$ .
- Write them in the brackets.
$x^2 + 7x + 12$ : two numbers multiplying to 12 and adding to 7 are 3 and 4, so $(x+3)(x+4)$ .
$x^2 - 2x - 15$ : multiplying to $-15$ , adding to $-2$ , gives $-5$ and $3$ , so $(x-5)(x+3)$ .
The sign rules that speed this up:
- $c$ positive — both numbers have the same sign, matching $b$
- $c$ negative — the numbers have opposite signs

When the leading coefficient is not 1

For $ax^2 + bx + c$ :
- Multiply $a$ by $c$ .
- Find two numbers multiplying to $ac$ and adding to $b$ .
- Split the middle term using them.
- Factorise in pairs.
Worked through for $6x^2 + 11x + 3$ :
- $ac = 18$ ; two numbers multiplying to 18 and adding to 11 are 9 and 2
- Split: $6x^2 + 9x + 2x + 3$
- Pair: $3x(2x + 3) + 1(2x + 3)$
- Factor out the common bracket: $(2x+3)(3x+1)$

The difference of two squares

$a^2 - b^2 = (a+b)(a-b)$

Only works for a subtraction. $a^2 + b^2$ does not factorise over the real numbers.
$9x^2 - 25 = (3x+5)(3x-5)$
$x^4 - 16 = \left(x^2+4\right)\left(x^2-4\right) = \left(x^2+4\right)(x+2)(x-2)$ — factorise fully, applying the rule again where possible.

Completing the square

Rewriting $x^2 + bx + c$ $x^{2} + b x + c$ as $(x + p)^2 + q$ $(x + p)^{2} + q$ :
- Halve the coefficient of $x$ to get $p$ .
- Square it and subtract it, then add the original constant.

$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c$

$x^2 + 6x + 11 = (x+3)^2 - 9 + 11 = (x+3)^2 + 2$
This form shows the vertex immediately: the minimum of $(x+3)^2 + 2$ is at $(-3, 2)$ , because a square is never negative.

Tips and Tricks

$(a+b)^2$ has three terms. Write out the four products the first few times rather than jumping to a memorised shortcut — the middle term is where the marks go.

Take out the common factor first, always. $2x^2 - 8 = 2(x^2 - 4) = 2(x+2)(x-2)$ is easy; trying to factorise $2x^2 - 8$ directly is not.

Factorise fully. Stopping at $\left(x^2+4\right)\left(x^2-4\right)$ when the second bracket still factorises loses the mark.

Worked ExampleFactorising fully

Factorise completely: (a) $3x^2 - 27$ , (b) $2x^2 - 7x - 15$ .

Part (a), Step 1 — Look for a common factor first

Both terms are divisible by 3:

$3x^2 - 27 = 3\left(x^2 - 9\right)$

Part (a), Step 2 — Recognise the difference of two squares

Inside the bracket, $x^2 - 9$ is $x^2 - 3^2$ , a difference of two squares:

$a^2 - b^2 = (a+b)(a-b) \quad \text{with } a = x, \; b = 3$

$= 3(x+3)(x-3)$

Part (a), Step 3 — Check by expanding

$3(x+3)(x-3) = 3\left(x^2 - 9\right) = 3x^2 - 27 \;\checkmark$

Part (b), Step 1 — Check for a common factor

$2$ , $7$ and $15$ share no common factor, so there is nothing to take out. The leading coefficient is not 1, so use the splitting method.

Part (b), Step 2 — Multiply $a$ by $c$

$a \times c = 2 \times (-15) = -30$

Part (b), Step 3 — Find two numbers

We need two numbers that multiply to $-30$ and add to $-7$ (the coefficient of $x$ ).

Since the product is negative, the two numbers have opposite signs. Testing pairs:

$-10$ and $3$ : product $-30$ ✓, sum $-7$ ✓

Part (b), Step 4 — Split the middle term

Replace $-7x$ with $-10x + 3x$ :

$2x^2 - 10x + 3x - 15$

Part (b), Step 5 — Factorise in pairs

Group the first two and the last two, and take a common factor from each:

$2x(x - 5) + 3(x - 5)$

Part (b), Step 6 — Take out the common bracket

$(x - 5)(2x + 3)$

Part (b), Step 7 — Check by expanding

$(x-5)(2x+3) = 2x^2 + 3x - 10x - 15 = 2x^2 - 7x - 15 \;\checkmark$

Expanding and factorising

Expanding brackets

Two brackets

Factorising — common factors

Factorising quadratics

When the leading coefficient is not 1

The difference of two squares

Completing the square

Part (a), Step 1 — Look for a common factor first

Part (a), Step 2 — Recognise the difference of two squares

Part (a), Step 3 — Check by expanding

Part (b), Step 1 — Check for a common factor

Part (b), Step 2 — Multiply aaa by ccc

Part (b), Step 3 — Find two numbers

Part (b), Step 4 — Split the middle term

Part (b), Step 5 — Factorise in pairs

Part (b), Step 6 — Take out the common bracket

Part (b), Step 7 — Check by expanding

Expanding and factorising

Expanding brackets

Two brackets

Factorising — common factors

Factorising quadratics

When the leading coefficient is not 1

The difference of two squares

Completing the square

Part (a), Step 1 — Look for a common factor first

Part (a), Step 2 — Recognise the difference of two squares

Part (a), Step 3 — Check by expanding

Part (b), Step 1 — Check for a common factor

Part (b), Step 2 — Multiply aaa by ccc

Part (b), Step 3 — Find two numbers

Part (b), Step 4 — Split the middle term

Part (b), Step 5 — Factorise in pairs

Part (b), Step 6 — Take out the common bracket

Part (b), Step 7 — Check by expanding

Part (b), Step 2 — Multiply $a$ by $c$

Part (b), Step 2 — Multiply $a$ by $c$