Factorising Harder Quadratics

What you'll learn

How to factorise quadratics where the coefficient of $x^2$ is not 1.
How to use the “split the middle term” method reliably.
How to solve quadratic equations once they are factorised.
How to handle difference of squares and quadratics involving $x$ and $y`.

Key words first

Definition

Important vocabulary

A quadratic expression has a highest power of 2, for example $2x^2 + 7x + 3$ .
A coefficient is the number multiplying a variable. In $ax^2 + bx + c$ , $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , and $c$ is the constant term.
Factorising means rewriting an expression as a product of brackets.
A binomial is an expression with two terms, such as $2 x + 1$ .

Warm-up: ordinary quadratics

Before harder quadratics, remember the simpler case where the coefficient of $x^2$ is 1.

For $x^2 + bx + c$ , you need two numbers that:

multiply to make $c$
add to make $b`

Example

Factorising a simple quadratic

Factorise $x^2 + 9x + 20$ .

An area model shows how the four parts combine to make x^2 + 9x + 20 and give the factors (x+4)(x+5).

Look for two numbers that multiply to 20 and add to 9.
The numbers are 4 and 5.
Write the brackets using those numbers:
$x^2 + 9x + 20 = (x + 4)(x + 5)$

Harder quadratics: when the front number is not 1

A harder quadratic has a coefficient in front of $x^2$ , such as $2x^2$ , $3x^2$ , or $5x^2`.

For these, the safest GCSE method is called splitting the middle term.

Key Idea

The split-the-middle method

For $ax^2 + bx + c$ , multiply $a$ and $c$ . Find two numbers that multiply to $a c$ and add to $b$ . Use those numbers to split the middle term.

The method

For $ax^2 + bx + c$ :

Multiply $a \times c`.
Find two numbers that multiply to $a c$ and add to $b`.
Split the $x$ term using those two numbers.
Factorise by grouping.
Take out the common bracket.

Example

Factorising a harder quadratic

Factorise $2x^2 + 7x + 3$ .

A factor grid makes the split 7x=6x+x visible before reading off the bracket factors.

Identify $a$ , $b$ , and $c`.
$2,\quad b = 7,\quad c = 3$
Multiply $a$ and $c`.
$\times 3 = 6$
Find two numbers that multiply to 6 and add to 7. They are 6 and 1.
Split the middle term:
$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$
Factorise each pair of terms:
$2 x (x + 3) + 1 (x + 3)$
Take out the common bracket:
$(2 x + 1) (x + 3)$

Tip

Quick check

Expand your answer mentally: first terms, outer terms, inner terms, last terms. The middle terms should combine to give the original middle term.

Handling negative signs

Signs are where many mistakes happen. The multiplying-and-adding rule still works, but you must include negative numbers.

If $c$ is positive and $b$ is negative, both split numbers are negative.
If $c$ is negative, one split number is positive and one is negative.

Example

Factorising with a negative constant

Factorise $4x^2 - 11x - 3$ .

The grid highlights that a negative constant needs one positive and one negative split term.

Multiply the first and last coefficients:
$\times (-3) = -12$
Find two numbers that multiply to -12 and add to -11. They are -12 and 1.
Split the middle term:
$4x^2 - 11x - 3 = 4x^2 - 12x + x - 3$
Factorise by grouping:
$4 x (x - 3) + 1 (x - 3)$
Take out the common bracket:
$(4 x + 1) (x - 3)$

Common Mistake

Ignoring the sign of the constant

If the constant term is negative, the two split numbers must have opposite signs. For example, multiplying to a negative number means one number is positive and one is negative.

Solving harder quadratic equations

An equation has an equals sign. To solve a quadratic equation by factorising, you first factorise, then use the fact that if two brackets multiply to zero, at least one bracket must be zero.

Definition

Zero product property

If $A B = 0$ , then $A = 0$ or $B = 0$ . This is why we set each bracket equal to zero after factorising.

Example

Solving a harder quadratic

Solve $3x^2 + 10x - 8 = 0$ .

A factor grid plus zero-product branches links factorising to solving the equation.

Multiply the first and last coefficients:
$\times (-8) = -24$
Find two numbers that multiply to -24 and add to 10. They are 12 and -2.
Split and factorise:
$3x2+10x−8=3x2+12x−2x−8=3x(x+4)−2(x+4)=(3x−2)(x+4)\begin{aligned} 3x^2 + 10x - 8 &= 3x^2 + 12x - 2x - 8 \\ &= 3x(x + 4) - 2(x + 4) \\ &= (3x - 2)(x + 4) \end{aligned}$
Set each bracket equal to zero:
$\quad \text{or} \quad x + 4 = 0$
Solve each linear equation:
$\frac{2}{3} \quad \text{or} \quad x = -4$

Factorising fully and difference of squares

Sometimes the first thing to do is take out a common factor: a number or variable that divides every term.

“Factorise fully” means keep factorising until you cannot factorise any further.

A useful identity is:

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

This is called the difference of two squares.

Example

Taking out a common factor first

Factorise fully $3x^2 - 75$ .

The diagram separates taking out the common factor from then using the difference of squares.

Take out the common factor of 3:
$3x^2 - 75 = 3(x^2 - 25)$
Notice that $x^2 - 25$ is a difference of squares:
$x^2 - 25 = x^2 - 5^2$
Factorise the difference of squares:
$3(x^2 - 25) = 3(x - 5)(x + 5)$

Tip

Common factor first

If every term shares a factor, take it out before using any other method. This makes the numbers smaller and avoids missing marks for not factorising fully.

Quadratics involving $x$ and $y$

Some quadratics contain terms like $x y$ and $y^2`. You can still use the same split-the-middle method.

Think of the middle term as the “ $x y$ term” instead of just the “ $x$ term”.

Example

Factorising a quadratic with two variables

Factorise $2x^2 + 11xy + 5y^2$ .

The two-variable grid treats the middle term as an xy term and reads off the factors in the same way.

Multiply the coefficient of $x^2$ by the coefficient of $y^2`:
$\times 5 = 10$
Find two numbers that multiply to 10 and add to 11. They are 10 and 1.
Split the middle term:
$2x^2 + 11xy + 5y^2 = 2x^2 + 10xy + xy + 5y^2$
Factorise by grouping:
$2 x (x + 5 y) + y (x + 5 y)$
Take out the common bracket:
$(2 x + y) (x + 5 y)$

Perfect-square quadratics

A perfect-square quadratic factorises into two identical brackets.

For example, $x^2 + 6xy + 9y^2$ has:

first term $x^2`
last term $(3y)^2`
middle term $2 \times x \times 3y`

Example

Spotting identical brackets

Factorise $x^2 - 8xy + 16y^2$ .

A square area model shows why the expression is a perfect square with two identical brackets.

Recognise the square terms:
$x2=x2,16y2=(4y)2x^2 = x^2,\quad 16y^2 = (4y)^2$
Check the middle term:
$\times x \times 4y = 8xy$
Because the middle term is negative, use minus signs:
$x^2 - 8xy + 16y^2 = (x - 4y)^2$

Exam technique

In the exam

First check for a common factor, especially when the question says “factorise fully”.
For $ax^2 + bx + c`, multiply$ a $an d$ c, then find two numbers that multiply to $ac$ and add to $b.
If the question says “solve”, do not stop at factorising: set each bracket equal to zero and find the values of $x`.

Self review

Check yourself

Can you explain why $2x^2 + 7x + 3$ splits into $2x^2 + 6x + x + 3$ ?
When the constant term is negative, what must be true about the signs of the two split numbers?
What is the first thing you should check before factorising fully?

Factorising Harder Quadratics

What you'll learn

How to factorise quadratics where the coefficient of $x^2$ is not 1.
How to use the “split the middle term” method reliably.
How to solve quadratic equations once they are factorised.
How to handle difference of squares and quadratics involving $x$ and $y`.

Key words first

Definition

Important vocabulary

A quadratic expression has a highest power of 2, for example $2x^2 + 7x + 3$ .
A coefficient is the number multiplying a variable. In $ax^2 + bx + c$ , $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ , and $c$ is the constant term.
Factorising means rewriting an expression as a product of brackets.
A binomial is an expression with two terms, such as $2 x + 1$ .

Warm-up: ordinary quadratics

Before harder quadratics, remember the simpler case where the coefficient of $x^2$ is 1.

For $x^2 + bx + c$ , you need two numbers that:

multiply to make $c$
add to make $b`

Example

Factorising a simple quadratic

Factorise $x^2 + 9x + 20$ .

An area model shows how the four parts combine to make x^2 + 9x + 20 and give the factors (x+4)(x+5).

Look for two numbers that multiply to 20 and add to 9.
The numbers are 4 and 5.
Write the brackets using those numbers:
$x^2 + 9x + 20 = (x + 4)(x + 5)$

Harder quadratics: when the front number is not 1

A harder quadratic has a coefficient in front of $x^2$ , such as $2x^2$ , $3x^2$ , or $5x^2`.

For these, the safest GCSE method is called splitting the middle term.

Key Idea

The split-the-middle method

For $ax^2 + bx + c$ , multiply $a$ and $c$ . Find two numbers that multiply to $a c$ and add to $b$ . Use those numbers to split the middle term.

The method

For $ax^2 + bx + c$ :

Multiply $a \times c`.
Find two numbers that multiply to $a c$ and add to $b`.
Split the $x$ term using those two numbers.
Factorise by grouping.
Take out the common bracket.

Example

Factorising a harder quadratic

Factorise $2x^2 + 7x + 3$ .

A factor grid makes the split 7x=6x+x visible before reading off the bracket factors.

Identify $a$ , $b$ , and $c`.
$2,\quad b = 7,\quad c = 3$
Multiply $a$ and $c`.
$\times 3 = 6$
Find two numbers that multiply to 6 and add to 7. They are 6 and 1.
Split the middle term:
$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$
Factorise each pair of terms:
$2 x (x + 3) + 1 (x + 3)$
Take out the common bracket:
$(2 x + 1) (x + 3)$

Tip

Quick check

Expand your answer mentally: first terms, outer terms, inner terms, last terms. The middle terms should combine to give the original middle term.

Handling negative signs

Signs are where many mistakes happen. The multiplying-and-adding rule still works, but you must include negative numbers.

If $c$ is positive and $b$ is negative, both split numbers are negative.
If $c$ is negative, one split number is positive and one is negative.

Example

Factorising with a negative constant

Factorise $4x^2 - 11x - 3$ .

The grid highlights that a negative constant needs one positive and one negative split term.

Multiply the first and last coefficients:
$\times (-3) = -12$
Find two numbers that multiply to -12 and add to -11. They are -12 and 1.
Split the middle term:
$4x^2 - 11x - 3 = 4x^2 - 12x + x - 3$
Factorise by grouping:
$4 x (x - 3) + 1 (x - 3)$
Take out the common bracket:
$(4 x + 1) (x - 3)$

Common Mistake

Ignoring the sign of the constant

If the constant term is negative, the two split numbers must have opposite signs. For example, multiplying to a negative number means one number is positive and one is negative.

Solving harder quadratic equations

An equation has an equals sign. To solve a quadratic equation by factorising, you first factorise, then use the fact that if two brackets multiply to zero, at least one bracket must be zero.

Definition

Zero product property

If $A B = 0$ , then $A = 0$ or $B = 0$ . This is why we set each bracket equal to zero after factorising.

Example

Solving a harder quadratic

Solve $3x^2 + 10x - 8 = 0$ .

A factor grid plus zero-product branches links factorising to solving the equation.

Multiply the first and last coefficients:
$\times (-8) = -24$
Find two numbers that multiply to -24 and add to 10. They are 12 and -2.
Split and factorise:
$3x2+10x−8=3x2+12x−2x−8=3x(x+4)−2(x+4)=(3x−2)(x+4)\begin{aligned} 3x^2 + 10x - 8 &= 3x^2 + 12x - 2x - 8 \\ &= 3x(x + 4) - 2(x + 4) \\ &= (3x - 2)(x + 4) \end{aligned}$
Set each bracket equal to zero:
$\quad \text{or} \quad x + 4 = 0$
Solve each linear equation:
$\frac{2}{3} \quad \text{or} \quad x = -4$

Factorising fully and difference of squares

Sometimes the first thing to do is take out a common factor: a number or variable that divides every term.

“Factorise fully” means keep factorising until you cannot factorise any further.

A useful identity is:

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

This is called the difference of two squares.

Example

Taking out a common factor first

Factorise fully $3x^2 - 75$ .

The diagram separates taking out the common factor from then using the difference of squares.

Take out the common factor of 3:
$3x^2 - 75 = 3(x^2 - 25)$
Notice that $x^2 - 25$ is a difference of squares:
$x^2 - 25 = x^2 - 5^2$
Factorise the difference of squares:
$3(x^2 - 25) = 3(x - 5)(x + 5)$

Tip

Common factor first

If every term shares a factor, take it out before using any other method. This makes the numbers smaller and avoids missing marks for not factorising fully.

Quadratics involving $x$ and $y$

Some quadratics contain terms like $x y$ and $y^2`. You can still use the same split-the-middle method.

Think of the middle term as the “ $x y$ term” instead of just the “ $x$ term”.

Example

Factorising a quadratic with two variables

Factorise $2x^2 + 11xy + 5y^2$ .

The two-variable grid treats the middle term as an xy term and reads off the factors in the same way.

Multiply the coefficient of $x^2$ by the coefficient of $y^2`:
$\times 5 = 10$
Find two numbers that multiply to 10 and add to 11. They are 10 and 1.
Split the middle term:
$2x^2 + 11xy + 5y^2 = 2x^2 + 10xy + xy + 5y^2$
Factorise by grouping:
$2 x (x + 5 y) + y (x + 5 y)$
Take out the common bracket:
$(2 x + y) (x + 5 y)$

Perfect-square quadratics

A perfect-square quadratic factorises into two identical brackets.

For example, $x^2 + 6xy + 9y^2$ has:

first term $x^2`
last term $(3y)^2`
middle term $2 \times x \times 3y`

Example

Spotting identical brackets

Factorise $x^2 - 8xy + 16y^2$ .

A square area model shows why the expression is a perfect square with two identical brackets.

Recognise the square terms:
$x2=x2,16y2=(4y)2x^2 = x^2,\quad 16y^2 = (4y)^2$
Check the middle term:
$\times x \times 4y = 8xy$
Because the middle term is negative, use minus signs:
$x^2 - 8xy + 16y^2 = (x - 4y)^2$

Exam technique

In the exam

First check for a common factor, especially when the question says “factorise fully”.
For $ax^2 + bx + c`, multiply$ a $an d$ c, then find two numbers that multiply to $ac$ and add to $b.
If the question says “solve”, do not stop at factorising: set each bracket equal to zero and find the values of $x`.

Self review

Check yourself

Can you explain why $2x^2 + 7x + 3$ splits into $2x^2 + 6x + x + 3$ ?
When the constant term is negative, what must be true about the signs of the two split numbers?
What is the first thing you should check before factorising fully?

Factorising Harder Quadratics

What you'll learn

Key words first

Warm-up: ordinary quadratics

Harder quadratics: when the front number is not 1

The method

Handling negative signs

Solving harder quadratic equations

Factorising fully and difference of squares

Quadratics involving $x$ and $y$

Perfect-square quadratics

Recap questions

You've reached the end

Factorising Harder Quadratics

What you'll learn

Key words first

Warm-up: ordinary quadratics

Harder quadratics: when the front number is not 1

The method

Handling negative signs

Solving harder quadratic equations

Factorising fully and difference of squares

Quadratics involving $x$ and $y$

Perfect-square quadratics

Recap questions

You've reached the end

Factorising Harder Quadratics

What you'll learn

Key words first

Warm-up: ordinary quadratics

Harder quadratics: when the front number is not 1

The method

Handling negative signs

Solving harder quadratic equations

Factorising fully and difference of squares

Quadratics involving xxx and yyy

Perfect-square quadratics

Recap questions

You've reached the end

Factorising Harder Quadratics Revision Guide

Factorising Harder Quadratics

What you'll learn

Key words first

Warm-up: ordinary quadratics

Harder quadratics: when the front number is not 1

The method

Handling negative signs

Solving harder quadratic equations

Factorising fully and difference of squares

Quadratics involving xxx and yyy

Perfect-square quadratics

Recap questions

You've reached the end

Factorising Harder Quadratics Revision Guide

Quadratics involving $x$ and $y$

Quadratics involving $x$ and $y$