Factorising Harder Quadratics

What you'll learn

How to factorise quadratics like $2x^2+7x+3$ , where the coefficient of $x^2$ is not 1.
How to choose the correct pair of numbers by using the product and sum.
How to solve quadratic equations after factorising.
How to handle special cases, including difference of two squares and quadratics with $x$ and $y$ .

Recap: what factorising means

Definition

Factorising

To factorise an expression means to rewrite it as a product of factors. For quadratics, this often means rewriting a three-term expression as two brackets multiplied together.

A quadratic is an expression where the highest power of the variable is 2, such as $x^2+5x+6$ or $3x^2-10x+8$ .

For easier quadratics like $x^2+7x+12$ , you look for two numbers that:

multiply to give the constant term, 12
add to give the coefficient of $x$ , which is 7

So the numbers are 3 and 4, giving:

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

Example

Recap: factorising when the coefficient of $x^{2}$ is 1

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

Look for two numbers that multiply to 20 and add to 9.
The pair is 4 and 5.
Write the expression as two brackets:
$x^2+9x+20=(x+4)(x+5)$

Key Idea

The easy case

When the quadratic starts with $x^2$ , the bracket beginnings are usually just $x$ and $x$ .

What makes a quadratic “harder”?

A harder quadratic has a coefficient in front of $x^2$ that is not 1, for example:

2x^2+7x+3

Here, the coefficient of $x^2$ is 2. That means the brackets may start with something like $2 x$ and $x$ , or $3 x$ and $2 x$ , depending on the expression.

Definition

Coefficient

A coefficient is the number multiplying a variable. In $5x^2-3x+8$ , the coefficient of $x^2$ is 5 and the coefficient of $x$ is -3.

The split-the-middle method

For a quadratic of the form $ax^2+bx+c$ , the reliable method is:

Multiply $a$ and $c$ .
Find two numbers that multiply to $a c$ and add to $b$ .
Split the middle term using those two numbers.
Factorise by grouping.

This method is sometimes called splitting the middle term.

Box method for factorising 6x^2 plus 17x plus 12

Example

Factorising a harder quadratic with positive terms

Factorise $3x^2+14x+8$ .

Identify $a$ , $b$ and $c$ :
$a=3,b=14,c=8a=3,\quad b=14,\quad c=8$
Multiply $a$ and $c$ :
$\times 8=24$
Find two numbers that multiply to 24 and add to 14. The pair is 12 and 2.
Split the middle term:
$3x^2+14x+8=3x^2+12x+2x+8$
Factorise the first pair and the second pair:
$3x^2+12x+2x+8=3x(x+4)+2(x+4)$
The repeated bracket is $x + 4$ , so factorise it out:
$3 x (x + 4) + 2 (x + 4) = (3 x + 2) (x + 4)$

Tip

The repeated bracket is your check

After grouping, both parts should contain the same bracket. If they do not, recheck your pair of numbers or your signs.

Handling negative signs

The signs are where many mistakes happen. Use the product and sum carefully.

For $ax^2+bx+c$ :

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

Example

Factorising when the constant term is negative

Factorise $2x^2+x-15$ .

Identify $a$ , $b$ and $c$ :
$a=2,b=1,c=−15a=2,\quad b=1,\quad c=-15$
Multiply $a$ and $c$ :
$\times (-15)=-30$
Find two numbers that multiply to -30 and add to 1. The pair is 6 and -5.
Split the middle term:
$2x^2+x-15=2x^2+6x-5x-15$
Factorise by grouping:
$2x^2+6x-5x-15=2x(x+3)-5(x+3)$
Take out the repeated bracket:
$2 x (x + 3) - 5 (x + 3) = (2 x - 5) (x + 3)$

Common Mistake

Forgetting the negative sign

If the product is negative, your two split numbers must have opposite signs. For example, to multiply to -30 and add to 1, you need 6 and -5, not 5 and -6.

Solving quadratic equations by factorising

Definition

Root

A root of an equation is a value of the variable that makes the equation true. For a factorised quadratic, roots come from making each bracket equal to zero.

Once a quadratic equation is factorised, use the zero product property:

\quad \text{means} \quad A=0 \text{ or } B=0

So if:

(2 x + 3) (x - 4) = 0

then either:

\quad \text{or} \quad x-4=0

Example

Solving a harder quadratic equation

Solve $4x^2+4x-15=0$ .

First factorise $4x^2+4x-15$ . Multiply $a$ and $c$ :
$\times (-15)=-60$
Find two numbers that multiply to -60 and add to 4. The pair is 10 and -6.
Split the middle term:
$4x^2+4x-15=4x^2+10x-6x-15$
Factorise by grouping:
$4x^2+10x-6x-15=2x(2x+5)-3(2x+5)$
Write the factorised form:
$4x^2+4x-15=(2x-3)(2x+5)$
Set each bracket equal to zero:
$\quad \text{or} \quad 2x+5=0$
Solve both linear equations:
$x=32orx=−52x=\frac{3}{2} \quad \text{or} \quad x=-\frac{5}{2}$

Tip

Factorise first, solve second

For a question that says “solve”, do not stop at the brackets. You need the values of $x$ .

Factorising fully

Sometimes a quadratic has a common factor first. Factorising fully means taking out every possible factor.

Example

Taking out a common factor first

Factorise fully $6x^2-24$ .

Look for the highest common factor of both terms. The highest common factor of 6 and 24 is 6.
Take out the common factor:
$6x^2-24=6(x^2-4)$
Notice that $x^2-4$ is a difference of two squares:
$x^2-4=x^2-2^2$
Factorise the difference of two squares:
$6(x^2-4)=6(x-2)(x+2)$

Definition

Difference of two squares

A difference of two squares has the form $a^2-b^2$ and factorises as $(a - b) (a + b)$ .

Common Mistake

Stopping too early

If you write $6(x^2-4)$ , you have factorised, but not fully. Because $x^2-4$ factorises again, the fully factorised answer is $6 (x - 2) (x + 2)$ .

Quadratics with $x$ and $y$

You may also see quadratics such as:

2x^2+7xy+3y^2

These work in a very similar way. The brackets will usually contain both $x$ and $y$ , for example:

(2 x + y) (x + 3 y)

When multiplying, remember:

\times y=xy,\quad y \times y=y^2

Example

Factorising a quadratic with two variables

Factorise $3x^2-13xy+4y^2$ .

Identify $a$ , $b$ and $c$ , treating $x y$ like the middle term:
$a=3,b=−13,c=4a=3,\quad b=-13,\quad c=4$
Multiply $a$ and $c$ :
$\times 4=12$
Find two numbers that multiply to 12 and add to -13. The pair is -12 and -1.
Split the middle term:
$3x^2-13xy+4y^2=3x^2-12xy-xy+4y^2$
Factorise by grouping:
$3x^2-12xy-xy+4y^2=3x(x-4y)-y(x-4y)$
Take out the repeated bracket:
$3 x (x - 4 y) - y (x - 4 y) = (3 x - y) (x - 4 y)$

Common Mistake

Watch the variable in the last term

If the last term is $4y^2$ , your brackets must contain $y$ terms, not just numbers. For example, $x - 4 y$ makes sense because multiplying $y$ by $y$ gives $y^2$ .

Quick checking by expanding

After factorising, you can check by expanding your brackets. This is especially useful in exams if your answer has awkward signs.

Example

Checking a factorisation

Check whether $(2 x - 3) (x + 5)$ is the factorisation of $2x^2+7x-15$ .

Expand the first terms:
$2x \times x=2x^2$
Expand the outside and inside terms:
$2x \times 5=10x,\quad -3 \times x=-3x$
Expand the last terms:
$\times 5=-15$
Combine the middle terms:
$2x^2+10x-3x-15=2x^2+7x-15$

Exam technique

In the exam

Multiply the first and last coefficients to find the product target, then find two numbers that add to the middle coefficient.
If the question says “solve”, factorise first and then set each bracket equal to zero.
Always check signs by expanding your brackets quickly, especially when the constant term is negative.

Self review

Check yourself

Can you factorise a quadratic like $2x^2+9x+10$ by splitting the middle term?
If $(3 x - 2) (x + 4) = 0$ , can you find both values of $x$ ?
Can you spot when a quadratic should be factorised further using difference of two squares?

Factorising Harder Quadratics

What you'll learn

How to factorise quadratics like $2x^2+7x+3$ , where the coefficient of $x^2$ is not 1.
How to choose the correct pair of numbers by using the product and sum.
How to solve quadratic equations after factorising.
How to handle special cases, including difference of two squares and quadratics with $x$ and $y$ .

Recap: what factorising means

Definition

Factorising

To factorise an expression means to rewrite it as a product of factors. For quadratics, this often means rewriting a three-term expression as two brackets multiplied together.

A quadratic is an expression where the highest power of the variable is 2, such as $x^2+5x+6$ or $3x^2-10x+8$ .

For easier quadratics like $x^2+7x+12$ , you look for two numbers that:

multiply to give the constant term, 12
add to give the coefficient of $x$ , which is 7

So the numbers are 3 and 4, giving:

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

Example

Recap: factorising when the coefficient of $x^{2}$ is 1

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

Look for two numbers that multiply to 20 and add to 9.
The pair is 4 and 5.
Write the expression as two brackets:
$x^2+9x+20=(x+4)(x+5)$

Key Idea

The easy case

When the quadratic starts with $x^2$ , the bracket beginnings are usually just $x$ and $x$ .

What makes a quadratic “harder”?

A harder quadratic has a coefficient in front of $x^2$ that is not 1, for example:

2x^2+7x+3

Here, the coefficient of $x^2$ is 2. That means the brackets may start with something like $2 x$ and $x$ , or $3 x$ and $2 x$ , depending on the expression.

Definition

Coefficient

A coefficient is the number multiplying a variable. In $5x^2-3x+8$ , the coefficient of $x^2$ is 5 and the coefficient of $x$ is -3.

The split-the-middle method

For a quadratic of the form $ax^2+bx+c$ , the reliable method is:

Multiply $a$ and $c$ .
Find two numbers that multiply to $a c$ and add to $b$ .
Split the middle term using those two numbers.
Factorise by grouping.

This method is sometimes called splitting the middle term.

Box method for factorising 6x^2 plus 17x plus 12

Example

Factorising a harder quadratic with positive terms

Factorise $3x^2+14x+8$ .

Identify $a$ , $b$ and $c$ :
$a=3,b=14,c=8a=3,\quad b=14,\quad c=8$
Multiply $a$ and $c$ :
$\times 8=24$
Find two numbers that multiply to 24 and add to 14. The pair is 12 and 2.
Split the middle term:
$3x^2+14x+8=3x^2+12x+2x+8$
Factorise the first pair and the second pair:
$3x^2+12x+2x+8=3x(x+4)+2(x+4)$
The repeated bracket is $x + 4$ , so factorise it out:
$3 x (x + 4) + 2 (x + 4) = (3 x + 2) (x + 4)$

Tip

The repeated bracket is your check

After grouping, both parts should contain the same bracket. If they do not, recheck your pair of numbers or your signs.

Handling negative signs

The signs are where many mistakes happen. Use the product and sum carefully.

For $ax^2+bx+c$ :

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

Example

Factorising when the constant term is negative

Factorise $2x^2+x-15$ .

Identify $a$ , $b$ and $c$ :
$a=2,b=1,c=−15a=2,\quad b=1,\quad c=-15$
Multiply $a$ and $c$ :
$\times (-15)=-30$
Find two numbers that multiply to -30 and add to 1. The pair is 6 and -5.
Split the middle term:
$2x^2+x-15=2x^2+6x-5x-15$
Factorise by grouping:
$2x^2+6x-5x-15=2x(x+3)-5(x+3)$
Take out the repeated bracket:
$2 x (x + 3) - 5 (x + 3) = (2 x - 5) (x + 3)$

Common Mistake

Forgetting the negative sign

If the product is negative, your two split numbers must have opposite signs. For example, to multiply to -30 and add to 1, you need 6 and -5, not 5 and -6.

Solving quadratic equations by factorising

Definition

Root

A root of an equation is a value of the variable that makes the equation true. For a factorised quadratic, roots come from making each bracket equal to zero.

Once a quadratic equation is factorised, use the zero product property:

\quad \text{means} \quad A=0 \text{ or } B=0

So if:

(2 x + 3) (x - 4) = 0

then either:

\quad \text{or} \quad x-4=0

Example

Solving a harder quadratic equation

Solve $4x^2+4x-15=0$ .

First factorise $4x^2+4x-15$ . Multiply $a$ and $c$ :
$\times (-15)=-60$
Find two numbers that multiply to -60 and add to 4. The pair is 10 and -6.
Split the middle term:
$4x^2+4x-15=4x^2+10x-6x-15$
Factorise by grouping:
$4x^2+10x-6x-15=2x(2x+5)-3(2x+5)$
Write the factorised form:
$4x^2+4x-15=(2x-3)(2x+5)$
Set each bracket equal to zero:
$\quad \text{or} \quad 2x+5=0$
Solve both linear equations:
$x=32orx=−52x=\frac{3}{2} \quad \text{or} \quad x=-\frac{5}{2}$

Tip

Factorise first, solve second

For a question that says “solve”, do not stop at the brackets. You need the values of $x$ .

Factorising fully

Sometimes a quadratic has a common factor first. Factorising fully means taking out every possible factor.

Example

Taking out a common factor first

Factorise fully $6x^2-24$ .

Look for the highest common factor of both terms. The highest common factor of 6 and 24 is 6.
Take out the common factor:
$6x^2-24=6(x^2-4)$
Notice that $x^2-4$ is a difference of two squares:
$x^2-4=x^2-2^2$
Factorise the difference of two squares:
$6(x^2-4)=6(x-2)(x+2)$

Definition

Difference of two squares

A difference of two squares has the form $a^2-b^2$ and factorises as $(a - b) (a + b)$ .

Common Mistake

Stopping too early

If you write $6(x^2-4)$ , you have factorised, but not fully. Because $x^2-4$ factorises again, the fully factorised answer is $6 (x - 2) (x + 2)$ .

Quadratics with $x$ and $y$

You may also see quadratics such as:

2x^2+7xy+3y^2

These work in a very similar way. The brackets will usually contain both $x$ and $y$ , for example:

(2 x + y) (x + 3 y)

When multiplying, remember:

\times y=xy,\quad y \times y=y^2

Example

Factorising a quadratic with two variables

Factorise $3x^2-13xy+4y^2$ .

Identify $a$ , $b$ and $c$ , treating $x y$ like the middle term:
$a=3,b=−13,c=4a=3,\quad b=-13,\quad c=4$
Multiply $a$ and $c$ :
$\times 4=12$
Find two numbers that multiply to 12 and add to -13. The pair is -12 and -1.
Split the middle term:
$3x^2-13xy+4y^2=3x^2-12xy-xy+4y^2$
Factorise by grouping:
$3x^2-12xy-xy+4y^2=3x(x-4y)-y(x-4y)$
Take out the repeated bracket:
$3 x (x - 4 y) - y (x - 4 y) = (3 x - y) (x - 4 y)$

Common Mistake

Watch the variable in the last term

If the last term is $4y^2$ , your brackets must contain $y$ terms, not just numbers. For example, $x - 4 y$ makes sense because multiplying $y$ by $y$ gives $y^2$ .

Quick checking by expanding

After factorising, you can check by expanding your brackets. This is especially useful in exams if your answer has awkward signs.

Example

Checking a factorisation

Check whether $(2 x - 3) (x + 5)$ is the factorisation of $2x^2+7x-15$ .

Expand the first terms:
$2x \times x=2x^2$
Expand the outside and inside terms:
$2x \times 5=10x,\quad -3 \times x=-3x$
Expand the last terms:
$\times 5=-15$
Combine the middle terms:
$2x^2+10x-3x-15=2x^2+7x-15$

Exam technique

In the exam

Multiply the first and last coefficients to find the product target, then find two numbers that add to the middle coefficient.
If the question says “solve”, factorise first and then set each bracket equal to zero.
Always check signs by expanding your brackets quickly, especially when the constant term is negative.

Self review

Check yourself

Can you factorise a quadratic like $2x^2+9x+10$ by splitting the middle term?
If $(3 x - 2) (x + 4) = 0$ , can you find both values of $x$ ?
Can you spot when a quadratic should be factorised further using difference of two squares?

Factorising Harder Quadratics

What you'll learn

Recap: what factorising means

What makes a quadratic “harder”?

The split-the-middle method

Handling negative signs

Solving quadratic equations by factorising

Factorising fully

Quadratics with $x$ and $y$

Quick checking by expanding

Recap questions

You've reached the end

Factorising Harder Quadratics

What you'll learn

Recap: what factorising means

What makes a quadratic “harder”?

The split-the-middle method

Handling negative signs

Solving quadratic equations by factorising

Factorising fully

Quadratics with $x$ and $y$

Quick checking by expanding

Recap questions

You've reached the end

Factorising Harder Quadratics

What you'll learn

Recap: what factorising means

What makes a quadratic “harder”?

The split-the-middle method

Handling negative signs

Solving quadratic equations by factorising

Factorising fully

Quadratics with xxx and yyy

Quick checking by expanding

Recap questions

You've reached the end

Factorising Harder Quadratics Revision Guide

Factorising Harder Quadratics

What you'll learn

Recap: what factorising means

What makes a quadratic “harder”?

The split-the-middle method

Handling negative signs

Solving quadratic equations by factorising

Factorising fully

Quadratics with xxx and yyy

Quick checking by expanding

Recap questions

You've reached the end

Factorising Harder Quadratics Revision Guide

Quadratics with $x$ and $y$

Quadratics with $x$ and $y$