## What is factorising equations?

Factorising equations describes the process of finding the factors. Essentially it is the reverse of expanding the brackets.

Factors are simply numbers that are multiplied together to get another number. An example of this is shown below:

Expressions can also have factors. An example of this is shown below:

## How to factorise an expression with 2 terms

There are two methods of factorising an expression with two terms.

**Method 1:**

Look for something in common between the 2 terms. Often this will be an “x” or “a” bracket (x+a)

Example 1a:

Factorise x² + 3x

In the example above, both terms have “x”, therefore = x(x+3)

Example 1b:

Factorise 3x(x-2) + 2(x-2)

In this example, the bracket “(x-2)” is common = (x-2) (3x+2)

**Method 2:**

The second method is where nothing is common. Usually, one term will be an ‘x²’ and the other term will be a number. This will be in the form a² – b². Recall that the difference of perfect squares a² – b² can be factorised into (a+b)(a+b).

Example 2a:

Factorise x² – 9

You have recognised that 9=3², and of course x² is a perfect square. We now use the difference of perfect square formula to factorise into (a+b)(a+b)

Therefore: x² – 9 = (x+√9) (x – √9)

= (x+3) (x-3)

Example 2b:

Factorise x² – 17

Even though 17 isn’t a perfect square, we can still factorise it using the difference of perfect squares formula. We just don’t get a nice number in the brackets.

x² – 17 = (x+√17) (x – √17)

Example 3:

Factorise 4px² – 256p

This is a combination of the two methods. Here, “4” is a common factor and “p” is common too, so the first step is to take out “4p”:

Therefore: 4p(x² – 64)

Now, recognise that inside the brackets we have a perfect square in the form a²-b². We now use the difference of perfect squares formula to factorise this further:

4p(x+8) (x-8)

## How to factorise an expression with 3 terms

These are expressions of the form ax² + bx + c. When the co-efficient of x² is 1, this is straight forward

Example 3a:

Factorise x² + 3x + 2

Step 1:

Find factors of the number (in this case the number is “2”).

The factors of 2 are “1” and “2”

Step 2: Using the factors, try and add or subtract to make the co-efficient of “x”, which is 3 in this case.

2+1=3

Step 3: Open two pairs of brackets:

( ) ( )

Step 4: Write “x” as the first term in each bracket:

(x )(x )

Step 5: We wanted +2 and +1 to make 3, so we write them in the brackets in any order:

(x+2) (x+1)

Therefore: x² + 3x + 2 = (x+2) (x+1)

Example 3b:

Factorise a² – 6a – 7

Step 1:

Factors of “7” are “1” and “7”

Step 2:

To make “-6”, we need -7 + 1

Step 3:

Open two pairs of brackets – ( ) ( )

Step 4:

Write “x” as the first term in each bracket – (x )(x )

Step 5:

Write in -7 and +1 – (x-7) (x+1)

Therefore: a² – 6a – 7 = (x-7) (x+1)

Example 3c:

Factorise t² – 6t + 8

Step 1:

Factors of “8” are “4”, “2”, “1” and “8”

Step 2:

To make “-6”, we need -4 and -2 (we can’t do anything with the 8 and 1 to make -6)

Step 3:

Open two pairs of brackets – ( ) ( )

Step 4:

Write “x” as the first term in each bracket – (x )(x )

Step 5:

Write in -4 and -2 – (x-4) (x-2)

Therefore: t² – 6t + 8 = (x-4) (x-2)

## Factorising a 3 term expression when the co-efficient is not x²

Example 4a:

Factorise 2x² + 5x + 2

Step 1: Here, we need to find factors of both the number “2” and the co-efficient of x² (also 2 in this case)

Step 2: Now, take one of the factors on the left and multiply it by one on the right. You do the same with the other pair:

Step 3: Open two pairs of brackets – ( ) ( )

Step 4: Instead of writing “x” as the first term, we have to take into account the factors we used. We split the “2” in front of the x² into 2×1 so we write 2x and 1x – (2x )(x )

Step 5: We paired the 2 with the other 2, and the 1 with the other 1. They go in **opposite** brackets – (2x 1)(x 2)

Step 6: The sign in whatever sign the factors of the number “2” were. In this case they were both positive

Therefore: 2x² + 5x + 2 = (2x+1) (x+2)

Example 4b:

Factorise 15x² + x – 2

Step 1: Find factors

Step 2: Trial and error:

15 × 2 = 30 1 × 1 = 1 ⇒ can’t make 1

15 × 1 = 15 1 × 2 = 2 ⇒ can’t make 1

5 × 2 = 10 3 × 1 = 3 ⇒ can’t make 1

5 × 1 = 5 3 × 2 = 6 ⇒ 6 – 5 = 1

So we want the 5 and 3 to go with the 2 and 1. The 5 goes with the 1 and the 3 with the 2.

(5x 2) (3x 1)

Step 3: We want to get positive 6 and negative 5. The 3×2 gave us 6, so both will be positive. The 5×1 gave us 5, so we need the 1 to be -1, so that we get -5.

Step 4: Fill in the signs

(5x+2) (3x-1)

Example 4c:

Factorise t² – 6t + 8

Step 1: Factors of “8” are “4”, “2”, “1” and “8”

Step 2: To make “-6”, we need -4 and -2 (we can’t do anything with the 8 and 1 to make -6)

Step 3: Open two pairs of brackets – ( ) ( )

Step 4: Write “x” as the first term in each bracket – (x )(x )

Step 5: Write in -4 and -2 – (x-4) (x-2)

Therefore: t² – 6t + 8 = (x-4) (x-2)

## Completing the square

In simple algebra, completing the square means to convert a quadratic equation by changing the form of the equation so the left side is a perfect square trinomial.

Example 5a:

Complete the square for x² + 10x

Step 1: Take the co-efficient of x (which is 10). Now divide it by 2 (which equals 5) and then square which results in 25.

Step 2: Add and subtract this number:

x² + 10x + 25 – 25

Step 3: The first 3 terms will factorise to give a perfect square.

x² + 10 + 25 – 25

(x+5) (x+5) – 25

(x+5)² – 25

Example 5b:

Complete the square for x² – 4x – 7

Step 1: Co-efficient of x is -4. Divide by 2 (which equals -2). Square this figure which results in 4.

Step 2: Add and subtract 4 (you can write the +4 then the -4).

x² – 4x +4 – 7 – 4

Step 3: The first 3 terms will factorise to give a perfect square.

x² – 4x + 4 – 7 – 4

= (x – 2) (x – 2) – 11

(x – 2)² – 11

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