Edexcel GCSE Mathematics · 1MA1

Expanding and Factorising Quadratics

What you need to know

Edexcel 1MA1 · A4 Expression Simplification Check the specification (PDF) (opens in a new tab)

Simplify and manipulate algebraic expressions, including those involving surds and algebraic fractions.
Collect like terms.
Multiply a single term over a bracket.
Take out common factors.
Expand products of two or more binomials.
Factorise quadratic expressions of the form $x^2 + bx + c$ , including the difference of two squares.
Factorise quadratic expressions of the form $ax^2 + bx + c$ .
Simplify expressions involving sums, products, and powers using the laws of indices.

Expanding and factorising are reverse processes

Brackets show multiplication. For example, $3(x+4)$ means that $3$ multiplies the whole expression $x+4$ . Expanding carries out the multiplication, while factorising reverses it:

$3(x+4)=3x+12$

Reading from left to right expands the expression; reading from right to left factorises it. This reverse relationship gives you a reliable check: expand any factorised answer and see whether you recover the original expression.

Expanding two binomials

A binomial contains two terms. When two binomials are multiplied, every term in the first bracket must multiply every term in the second. For example,

$(x+3)(x+5)$

produces four products:

$x\times x+x\times5+3\times x+3\times5$

$(x+3)(x+5)=x^2+5x+3x+15=x^2+8x+15.$

The terms $5x$ and $3x$ are like terms, so they combine to make $8x$ . A grid can organise the same four products:

$\times$	$x$	$5$
$x$	$x^2$	$5x$
$3$	$3x$	$15$

Add all four entries and collect like terms.

Keeping control of signs

A sign belongs to the term immediately after it. In $(x+7)(x-5)$ , the second term is $-5$ , not $5$ . Therefore

$(x+7)(x-5)=x^2-5x+7x-35=x^2+2x-35.$

The same principle applies when both brackets contain negative terms:

$(2x-3)(x-4)=2x^2-8x-3x+12=2x^2-11x+12.$

The constant is positive because $(-3)(-4)=12$ .

A squared binomial means two identical factors. Thus

$(x-6)^2=(x-6)(x-6)=x^2-12x+36.$

Both cross-products are $-6x$ , which is why the middle term is $-12x$ .

Products of three binomials

This course can assess expanded products of up to three binomials, producing a cubic expression. Do not try to multiply all three simultaneously. Expand two brackets first, retain brackets around that result, and then multiply by the third.

For example,

$(x-1)(x+2)(x+3)$

First expand two brackets:

$(x-1)(x+2)=x^2+x-2.$

Then multiply every term of this result by both terms in the remaining bracket:

\begin{aligned} (x^2+x-2)(x+3) &=x^3+x^2-2x+3x^2+3x-6\\ &=x^3+4x^2+x-6. \end{aligned}

Substitution offers a useful check. At $x=1$ , the original product is $0\times3\times4=0$ , while the expansion gives $1+4+1-6=0$ .

Factorising $x^2+bx+c$

Factorising a monic quadratic reverses the expansion

$(x+p)(x+q)=x^2+(p+q)x+pq.$

Therefore, to factorise $x^2+bx+c$ , find two numbers whose:

sum is $b$ ;
product is $c$ .

For $x^2+9x+20$ , the numbers are $4$ and $5$ because $4+5=9$ and $4\times5=20$ . Hence

$x^2+9x+20=(x+4)(x+5).$

For $x^2-2x-15$ , the product is negative, so the two numbers have opposite signs. The required numbers are $3$ and $-5$ :

$3+(-5)=-2,\qquad 3(-5)=-15,$

and therefore

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

When $c$ is positive, the two numbers have the same sign. Their shared sign is determined by $b$ : for example,

$x^2-7x+12=(x-3)(x-4).$

The difference of two squares

Expanding $(A+B)(A-B)$ gives

$A^2-AB+AB-B^2=A^2-B^2.$

The middle terms cancel, producing the identity

$A^2-B^2=(A+B)(A-B).$

This applies whenever both terms are perfect squares and they are separated by subtraction:

$x^2-49=x^2-7^2=(x+7)(x-7),$

and

$9x^2-25=(3x)^2-5^2=(3x+5)(3x-5).$

It does not directly apply to a sum such as $x^2+49$ .

Factorising $ax^2+bx+c$

When the coefficient of $x^2$ is not $1$ , the factors may begin with different multiples of $x$ . One systematic method is to multiply $a$ and $c$ , then split the middle term.

Consider

$6x^2+11x+3.$

Here $a\times c=6\times3=18$ . Find two numbers with product $18$ and sum $11$ : they are $9$ and $2$ . Split $11x$ using those numbers:

$6x^2+9x+2x+3.$

Factorise the first pair and the second pair:

$3x(2x+3)+1(2x+3).$

The repeated bracket is now a common factor:

$6x^2+11x+3=(3x+1)(2x+3).$

For a quadratic with a negative constant, the splitting numbers have opposite signs. For example, in $2x^2-7x-15$ , $a\times c=-30$ . The numbers $-10$ and $3$ multiply to $-30$ and add to $-7$ :

\begin{aligned} 2x^2-7x-15 &=2x^2-10x+3x-15\\ &=2x(x-5)+3(x-5)\\ &=(2x+3)(x-5). \end{aligned}

Expanding the final brackets confirms the factorisation.

Factorise fully

Always look for a common factor before using a quadratic method. For example,

$3x^2+12x+9=3(x^2+4x+3)=3(x+1)(x+3).$

Stopping at $3(x^2+4x+3)$ would not fully factorise the quadratic. The same idea can reveal a difference of two squares:

$2x^2-32=2(x^2-16)=2(x+4)(x-4).$

The completed answer is a product, and no factor inside it can be factorised further using the methods above.

Expand

Multiply every term in one bracket by every term in the other, then collect like terms:

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

A square means identical brackets:

$(x+p)^2=x^2+2px+p^2,$

$(x-p)^2=x^2-2px+p^2.$

For three binomials, expand two first and multiply the resulting expression by the third. Assessed expansions are limited to products of three binomials.

Factorise $x^2+bx+c$

Find $p$ and $q$ such that

$p+q=b,\qquad pq=c.$

Then

$x^2+bx+c=(x+p)(x+q).$

Sign recall:

$c<0$ : opposite signs;
$c>0$ , $b>0$ : both positive;
$c>0$ , $b<0$ : both negative.

Difference of two squares

$A^2-B^2=(A+B)(A-B).$

Both terms must be squares and the operation between them must be subtraction.

Factorise $ax^2+bx+c$

Calculate $ac$ .
Find two numbers with product $ac$ and sum $b$ .
Split the middle term using those numbers.
Factorise by grouping.

Final checks

Take out any common factor first, factorise completely, and expand your answer to verify it reproduces the original quadratic.

Exam tips

1
For “expand and simplify”, multiply every term required and then collect like terms; an uncollected expansion is not fully simplified.
2
Treat a negative term as a complete signed term. In a grid, write $-5$ , not just $5$ .
3
When squaring a binomial, write it as two identical factors: $(x-4)^2=(x-4)(x-4)$ . It is not $x^2+16$ .
4
When expanding three binomials, expand two first, keep their result in brackets, and then multiply by the third.
5
Before factorising a quadratic, check whether all its terms have a common factor. Take out the highest common factor first.
6
After factorising, expand the brackets mentally or in writing. The first terms, middle terms and constant term must reproduce the original expression.
7
For a difference of two squares, check that there are exactly two square terms separated by subtraction: $A^2-B^2=(A+B)(A-B)$ .
8
A factorisation should normally be complete: no further common factor should remain inside a bracket.

Ask the tutor about this note

It can see this note and what your specification asks for.

Answers are free with an account.

1 free note left this week

Get more free notes

How well do you know this now?

Your answer sets when this note comes back for review, or marks it complete.

Was this note helpful?

Last updated 5 Oct 2026

Key terms(8)

Expand: To multiply out brackets so that an equivalent expression is written without those brackets.
Factorise: To rewrite an expression as a product of factors. It is the reverse of expanding.
Factor: A number, variable or algebraic expression that multiplies another quantity.
Like terms: Terms that have exactly the same variable part, including the same powers, and can therefore be collected.
Quadratic expression: An expression whose highest power of the variable is 2, such as $3x^2-8x+4$ .
Binomial: An expression containing two terms, such as $x+5$ or $2x-3$ .
Difference of two squares: A pattern of the form $A^2-B^2$ , which factorises as $(A+B)(A-B)$ .
Common factor: A factor shared by every term in an expression.

Previous noteBasic Algebraic Manipulation and Indices Next noteSurds and Algebraic Fractions

Back to Expression Simplification

Put your knowledge into practice — try past paper questions for Mathematics

Edexcel GCSE Mathematics · 1MA1

Expanding and Factorising Quadratics

What you need to know

Edexcel 1MA1 · A4 Expression Simplification Check the specification (PDF) (opens in a new tab)

Simplify and manipulate algebraic expressions, including those involving surds and algebraic fractions.
Collect like terms.
Multiply a single term over a bracket.
Take out common factors.
Expand products of two or more binomials.
Factorise quadratic expressions of the form $x^2 + bx + c$ , including the difference of two squares.
Factorise quadratic expressions of the form $ax^2 + bx + c$ .
Simplify expressions involving sums, products, and powers using the laws of indices.

Expanding and factorising are reverse processes

Brackets show multiplication. For example, $3(x+4)$ means that $3$ multiplies the whole expression $x+4$ . Expanding carries out the multiplication, while factorising reverses it:

$3(x+4)=3x+12$

Expanding two binomials

A binomial contains two terms. When two binomials are multiplied, every term in the first bracket must multiply every term in the second. For example,

$(x+3)(x+5)$

produces four products:

$x\times x+x\times5+3\times x+3\times5$

$(x+3)(x+5)=x^2+5x+3x+15=x^2+8x+15.$

The terms $5x$ and $3x$ are like terms, so they combine to make $8x$ . A grid can organise the same four products:

$\times$	$x$	$5$
$x$	$x^2$	$5x$
$3$	$3x$	$15$

Add all four entries and collect like terms.

Keeping control of signs

A sign belongs to the term immediately after it. In $(x+7)(x-5)$ , the second term is $-5$ , not $5$ . Therefore

$(x+7)(x-5)=x^2-5x+7x-35=x^2+2x-35.$

The same principle applies when both brackets contain negative terms:

$(2x-3)(x-4)=2x^2-8x-3x+12=2x^2-11x+12.$

The constant is positive because $(-3)(-4)=12$ .

A squared binomial means two identical factors. Thus

$(x-6)^2=(x-6)(x-6)=x^2-12x+36.$

Both cross-products are $-6x$ , which is why the middle term is $-12x$ .

Products of three binomials

For example,

$(x-1)(x+2)(x+3)$

First expand two brackets:

$(x-1)(x+2)=x^2+x-2.$

Then multiply every term of this result by both terms in the remaining bracket:

\begin{aligned} (x^2+x-2)(x+3) &=x^3+x^2-2x+3x^2+3x-6\\ &=x^3+4x^2+x-6. \end{aligned}

Substitution offers a useful check. At $x=1$ , the original product is $0\times3\times4=0$ , while the expansion gives $1+4+1-6=0$ .

Factorising $x^2+bx+c$

Factorising a monic quadratic reverses the expansion

$(x+p)(x+q)=x^2+(p+q)x+pq.$

Therefore, to factorise $x^2+bx+c$ , find two numbers whose:

sum is $b$ ;
product is $c$ .

For $x^2+9x+20$ , the numbers are $4$ and $5$ because $4+5=9$ and $4\times5=20$ . Hence

$x^2+9x+20=(x+4)(x+5).$

For $x^2-2x-15$ , the product is negative, so the two numbers have opposite signs. The required numbers are $3$ and $-5$ :

$3+(-5)=-2,\qquad 3(-5)=-15,$

and therefore

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

When $c$ is positive, the two numbers have the same sign. Their shared sign is determined by $b$ : for example,

$x^2-7x+12=(x-3)(x-4).$

The difference of two squares

Expanding $(A+B)(A-B)$ gives

$A^2-AB+AB-B^2=A^2-B^2.$

The middle terms cancel, producing the identity

$A^2-B^2=(A+B)(A-B).$

This applies whenever both terms are perfect squares and they are separated by subtraction:

$x^2-49=x^2-7^2=(x+7)(x-7),$

and

$9x^2-25=(3x)^2-5^2=(3x+5)(3x-5).$

It does not directly apply to a sum such as $x^2+49$ .

Factorising $ax^2+bx+c$

When the coefficient of $x^2$ is not $1$ , the factors may begin with different multiples of $x$ . One systematic method is to multiply $a$ and $c$ , then split the middle term.

Consider

$6x^2+11x+3.$

Here $a\times c=6\times3=18$ . Find two numbers with product $18$ and sum $11$ : they are $9$ and $2$ . Split $11x$ using those numbers:

$6x^2+9x+2x+3.$

Factorise the first pair and the second pair:

$3x(2x+3)+1(2x+3).$

The repeated bracket is now a common factor:

$6x^2+11x+3=(3x+1)(2x+3).$

For a quadratic with a negative constant, the splitting numbers have opposite signs. For example, in $2x^2-7x-15$ , $a\times c=-30$ . The numbers $-10$ and $3$ multiply to $-30$ and add to $-7$ :

\begin{aligned} 2x^2-7x-15 &=2x^2-10x+3x-15\\ &=2x(x-5)+3(x-5)\\ &=(2x+3)(x-5). \end{aligned}

Expanding the final brackets confirms the factorisation.

Factorise fully

Always look for a common factor before using a quadratic method. For example,

$3x^2+12x+9=3(x^2+4x+3)=3(x+1)(x+3).$

Stopping at $3(x^2+4x+3)$ would not fully factorise the quadratic. The same idea can reveal a difference of two squares:

$2x^2-32=2(x^2-16)=2(x+4)(x-4).$

The completed answer is a product, and no factor inside it can be factorised further using the methods above.

Expand

Multiply every term in one bracket by every term in the other, then collect like terms:

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

A square means identical brackets:

$(x+p)^2=x^2+2px+p^2,$

$(x-p)^2=x^2-2px+p^2.$

For three binomials, expand two first and multiply the resulting expression by the third. Assessed expansions are limited to products of three binomials.

Factorise $x^2+bx+c$

Find $p$ and $q$ such that

$p+q=b,\qquad pq=c.$

Then

$x^2+bx+c=(x+p)(x+q).$

Sign recall:

$c<0$ : opposite signs;
$c>0$ , $b>0$ : both positive;
$c>0$ , $b<0$ : both negative.

Difference of two squares

$A^2-B^2=(A+B)(A-B).$

Both terms must be squares and the operation between them must be subtraction.

Factorise $ax^2+bx+c$

Calculate $ac$ .
Find two numbers with product $ac$ and sum $b$ .
Split the middle term using those numbers.
Factorise by grouping.

Final checks

Take out any common factor first, factorise completely, and expand your answer to verify it reproduces the original quadratic.

Exam tips

1
For “expand and simplify”, multiply every term required and then collect like terms; an uncollected expansion is not fully simplified.
2
Treat a negative term as a complete signed term. In a grid, write $-5$ , not just $5$ .
3
When squaring a binomial, write it as two identical factors: $(x-4)^2=(x-4)(x-4)$ . It is not $x^2+16$ .
4
When expanding three binomials, expand two first, keep their result in brackets, and then multiply by the third.
5
Before factorising a quadratic, check whether all its terms have a common factor. Take out the highest common factor first.
6
After factorising, expand the brackets mentally or in writing. The first terms, middle terms and constant term must reproduce the original expression.
7
For a difference of two squares, check that there are exactly two square terms separated by subtraction: $A^2-B^2=(A+B)(A-B)$ .
8
A factorisation should normally be complete: no further common factor should remain inside a bracket.

Ask the tutor about this note

It can see this note and what your specification asks for.

Answers are free with an account.

1 free note left this week

Get more free notes

How well do you know this now?

Your answer sets when this note comes back for review, or marks it complete.

Was this note helpful?

Last updated 5 Oct 2026

Key terms(8)

Expand: To multiply out brackets so that an equivalent expression is written without those brackets.
Factorise: To rewrite an expression as a product of factors. It is the reverse of expanding.
Factor: A number, variable or algebraic expression that multiplies another quantity.
Like terms: Terms that have exactly the same variable part, including the same powers, and can therefore be collected.
Quadratic expression: An expression whose highest power of the variable is 2, such as $3x^2-8x+4$ .
Binomial: An expression containing two terms, such as $x+5$ or $2x-3$ .
Difference of two squares: A pattern of the form $A^2-B^2$ , which factorises as $(A+B)(A-B)$ .
Common factor: A factor shared by every term in an expression.

Previous noteBasic Algebraic Manipulation and Indices Next noteSurds and Algebraic Fractions

Back to Expression Simplification

Put your knowledge into practice — try past paper questions for Mathematics

What you need to know

Expanding and factorising are reverse processes

Expanding two binomials

Keeping control of signs

Products of three binomials

Factorising x2+bx+cx^2+bx+cx2+bx+c

The difference of two squares

Factorising ax2+bx+cax^2+bx+cax2+bx+c

Factorise fully

Expand

Factorise x2+bx+cx^2+bx+cx2+bx+c

Difference of two squares

Factorise ax2+bx+cax^2+bx+cax2+bx+c

Final checks

Exam tips

Ask the tutor about this note

How well do you know this now?

Key terms(8)

Key terms(8)

What you need to know

Expanding and factorising are reverse processes

Expanding two binomials

Keeping control of signs

Products of three binomials

Factorising x2+bx+cx^2+bx+cx2+bx+c

The difference of two squares

Factorising ax2+bx+cax^2+bx+cax2+bx+c

Factorise fully

Expand

Factorise x2+bx+cx^2+bx+cx2+bx+c

Difference of two squares

Factorise ax2+bx+cax^2+bx+cax2+bx+c

Final checks

Exam tips

Ask the tutor about this note

How well do you know this now?

Key terms(8)

Key terms(8)

Factorising $x^2+bx+c$

Factorising $ax^2+bx+c$

Factorise $x^2+bx+c$

Factorise $ax^2+bx+c$

Factorising $x^2+bx+c$

Factorising $ax^2+bx+c$

Factorise $x^2+bx+c$

Factorise $ax^2+bx+c$