# Factorising Quadratics

## Factorising Quadratics Revision

**Factorising Quadratics**

**Quadratics** are expressions that, in their most general form, look like

ax^{2}+bx+c

where a, b and c are all numbers, called **coefficients**.

Some **quadratics** **factorise** into **double brackets**, which take the form as follows:

ax^{2}+bx+c=(px+q)(rx+s)

where p, q, r and s are all numbers.

There are different rules on how to **factorise** **quadratics**, depending on if a=1 or a\neq1

**Note: Positives and Negatives**

Assuming a>0, you can easily determine whether each **bracket** will contain a + or a - by finding which of the three cases below your **quadratic** falls into.

**Case 1:** b>0 and c>0

In this case, **both brackets** will contain a + sign.

**Example:** x^{2}+10x+24=(x+6)(x+4)

In this example, b=10>0 and c=24>0, so **both brackets** contain a + sign.

**Case 2:** b<0 and c>0

In this case, **both brackets** will contain a - sign.

**Example:** x^{2}-6x+8=(x-4)(x-2)

In this example, b=-6<0 and c=8>0, so **both brackets** contain a - sign.

**Case 3:** c<0

In this case, regardless of the value of b, **one bracket** will contain a + sign and the **other bracket** will contain a - sign.

**Example:** x^{2}-x-6=(x+2)(x-3)

In this example, c=-6<0 so **one bracket** contains a + sign and the **other bracket** contains a - sign.

**Factorising Quadratics (a=1)**

**Quadratics** where a=1 are **quadratics** of the form:

x^{2}+bx+c

The **brackets** will be of the form (x+p)(x+q) where p and q **add** to give b and **multiply** to give c. In other words:

p+q=b

pq=c

**Example:** **Factorise** x^{2}+4x-12

We need two numbers that **add** to give 4 and **multiply** to give -12.

We know that **one of the numbers must be positive and one of the numbers must be negative**, because c=-12 so c<0.

\textcolor{blue}{-2}+\textcolor{red}{6}=4

\textcolor{blue}{-2}\times\textcolor{red}{6}=12

So the numbers must be \textcolor{blue}{-2} and \textcolor{red}{6}.

Putting it all together, we have:

x^{2}+4x-12=(x-2)(x+6)

**Factorising Quadratics (a\neq1)**

When a\neq1, there will be **coefficients** in front of the x terms inside the **brackets**, i.e. the **factorisation** will take the form:

ax^{2}+bx+c=(px+q)(rx+s)

From this, we know that a=pr and c=qs, so we can generate **many possible pairs** for these numbers. However, the only way to find the right set is to try them all and see which gives the correct value for b.

**Example:** **Factorise** 3x^{2}+13x+12

Firstly, since **all terms are positive**, we know all of p, q, r and s are going to be **positive**.

Next, the only pair of numbers that multiply to give 3 are 3 and 1, so these must be p and r

Next, to get 12, we have lots of options:

1\times12=12

2\times6=12

3\times4=12

One of these pairs will be q and s.

Next, start **expanding every possible factorisation combination**, until the correct b value is found.

(3x+12)(x+1)=3x^{2}+15x+12 has b=15\neq13

(x+12)(3x+1)=3x^{2}+37x+12 has b=37\neq13

(3x+6)(x+2)=3x^{2}+12x+12 has b=12\neq13

(x+6)(3x+2)=3x^{2}+20x+12 has b=20\neq13

(3x+4)(x+3)=3x^{2}+13x+12 has b=13 is the correct factorisation.

At this point, even though we have one combination left to try, we can stop because **we have found the correct set of brackets**.

Hence, we can conclude that 3x^{2}+13x+12=(3x+4)(x+3)

**Example 1: Factorising Quadratics (a=1)**

**Factorise** x^{2}+3x+2

**[2 marks]**

Since b=3>0 and c=2>0, both **brackets** will contain a + sign.

We need to find two numbers that **add** to give 3 and **multiply** to give 2.

These numbers are 1 and 2, so 1 and 2 are the numbers in the brackets.

Hence, our **factorised** **quadratic** is:

x^{2}+3x+2=(x+1)(x+2)

**Example 2: Factorising Quadratics (a\neq1)**

**Factorise** 10x^{2}-3x-1

**[3 marks]**

Since a\neq1, we know that this **factorisation** must take the form (px+q)(rx+s).

We also know that pr=10 and qs=-1.

There are **multiple pairs** for p abnd r:

10\times1=10

5\times2=10

For q and s, there is only one pair: 1 and -1.

Try every combination:

(10x+1)(x-1)=10x^{2}-9x-1

(10x-1)(x+1)=10x^{2}+9x-1

(5x+1)(2x-1)=10x^{2}-3x-1

We can stop here because **we have found the correct value** for b.

Hence, we can conclude that:

10x^{2}-3x-1=(5x+1)(2x-1)

## Factorising Quadratics Example Questions

**Question 1:** Factorise y^{2}-2y+1

**[2 marks]**

We need two numbers that add to give -2 and multiply to give 1.

Since b=-2<0 and c=1>0, we know that both numbers will be negative.

The numbers are -1 and -1.

Hence,

y^{2}-2y+1=(y-1)(y-1)=(y-1)^{2}**Question 2:** Factorise x^{2}+4x-21

**[2 marks]**

We need two numbers that add to give 4 and multiply to give -21.

Since c=-21<0, we know that one number will be positive and the other will be negative.

The possible pairs to multiply to -21 are:

1 and -21

-1 and 21

3 and -7

-3 and 7

The correct pair is -3 and 7, because these add to give 4

Hence,

x^{2}+4x-21=(x-3)(x+7)**Question 3:** Factorise x^{2}+52x+147

**[2 marks]**

We need two numbers that add to give 52 and multiply to give 147.

Since b=52>0 and c=147>0, we know that both numbers will be positive.

The possible pairs to multiply to 147 are:

1 and 147

3 and 49

7 and 21

The correct pair is 3 and 49, because these add to give 52

Hence,

x^{2}+52x+147=(x+3)(x+49)**Question 4:** Factorise 2x^{2}+13x+11

**[3 marks]**

We know that this will take the form (px+q)(rx+s).

We also know that, since b=13>0 and c=11>0, all of p, q, r and s must be positive.

pr=2 – the only pair that makes 2 is 1 and 2.

qs=11 – the only pair that makes 11 is 1 and 11.

This gives only two possibilites: (2x+1)(x+11) and (2x+11)(x+1)

Multiply both out to find correct answer:

(2x+1)(x+11)=2x^{2}+23x+11 has b=23\neq13

(2x+11)(x+1)=2x^{2}+13x+11 has b=13 so must be the correct answer.

Hence,

2x^{2}+13x+11=(2x+11)(x+1)**Question 5:** Factorise 12x^{2}-29x-221

**[3 marks]**

We know this will take the form (px+q)(rx+s)

Since c=-221<0, we know that p>0, r>0 and one of q and s is positive while the other is negative.

pr=12, the possible pairs for which are:

1 and 12

2 and 6

3 and 4

qs=-221, the possible pairs for which are:

1 and -221

-1 and 221

13 and -17

-13 and 17

Now start trying combinations:

(x+1)(12x-221)=12x^{2}-209x-221

(x-1)(12x+221)=12x^{2}+209x-221

(x+13)(12x-17)=12x^{2}+139x-221

(x-13)(12x+17)=12x^{2}-139x-221

(2x+1)(6x-221)=12x^{2}-436x-221

(2x-1)(6x+221)=12x^{2}+436x-221

(2x+13)(6x-17)=12x^{2}+42x-221

(2x-13)(6x+17)=12x^{2}-42x-221

(4x+1)(3x-221)=12x^{2}-881x-221

(4x-1)(3x+221)=12x^{2}+881x-221

(4x+13)(3x-17)=12x^{2}-29x-221

Although there are several other combinations to try, we can stop here because we have found the correct b value, so we know this must be the correct factorisation.

Hence,

12x^{2}-29x-221=(4x+13)(3x-17)## You May Also Like...

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