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WJEC MATHEMATICS
INTERMEDIATE
ALGEBRA
FACTORISING
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Contents
Highest Common Factors (algebra)
Factorising to a single bracket
Factorising to double brackets
Credits
WJEC Question bank
http://www.wjec.co.uk/question-bank/question-search.html
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In order to factorise effectively, make sure you understand the term
'highest common factor' of two numbers. You can find this in the
booklet 'Number properties'
Highest Common Factor (algebraic)
We know how to find the HCF of two numbers (e.g. 18 and 20) but
what about when algebra is involved? To find this, we need to use
our knowledge of indices to break the terms down.
Example
Find the HCF of 𝑥 4 𝑎𝑛𝑑 𝑥 6
Let's 'break down' each term
𝑥 4 = 𝑥×𝑥×𝑥×𝑥
𝑥 6 = 𝑥×𝑥×𝑥×𝑥×𝑥×𝑥
Now, we can see what appears in both lists. In both lists we can see
𝑥×𝑥×𝑥×𝑥 so 𝑥 4 is the HCF.
Let's try something more difficult
Example 2
Find the HCF of 𝑥 4 𝑦 3 and 𝑥 2 𝑦 5
Let's 'break down' each term
𝑥 4 𝑦 3 = 𝑥×𝑥×𝑥×𝑥×𝑦×𝑦×𝑦
𝑥 2 𝑦 5 = 𝑥×𝑥×𝑦×𝑦×𝑦×𝑦×𝑦
Now we can see what appears in both lists. We can see the HCF of
these terms is 𝑥×𝑥×𝑦×𝑦×𝑦 = 𝑥 2 𝑦 3
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When finding the HCF of two terms that contain numbers and
algebra, consider the numbers and algebra separately.
Example 3
Find the HCF of 10𝑎𝑏 and 25𝑎
The HCF of 10 and 25 = 5
The HCF of 𝑎𝑏 and 𝑎 = 𝑎
So, the HCF of 10𝑎𝑏 and 25𝑎 = 5𝑎
Exercise A4
Complete the table
Term 1
𝒙×𝑥×𝑥×𝒚
𝑥 3𝑦
𝑥2
𝑥3
𝑎6
𝑎4𝑏
𝑥 3𝑦 2
𝑎2 𝑏2
6𝑥 3
8𝑦 4
7𝑥 2
16𝑥 3 𝑦
Term 2
𝑥𝑦 2
𝑥4
𝑥
𝑎8
𝑎𝑏
𝑥 2𝑦 3
𝑎𝑏
4𝑥 2
10𝑦 6
𝑥
24𝑥𝑦 3
HCF
𝒙×𝑦×𝒚
𝑥𝑦
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When factorising, we need to introduce brackets. It is the opposite of
expanding. A good knowledge of expanding will make this section
easier and allow you to check your answers.
Expanding to a Single Bracket
When you are given two terms to factorise, there are three steps;
Example
Factorise
12𝑥 2 + 8𝑥
Step 1: Find the HCF of both terms and put this outside the brackets
The HCF of 12𝑥 2 and 8𝑥 is 4𝑥
= 4𝑥(
+
)
Step 2: The HCF multiplied by the first term in the bracket needs to
give us 12𝑥 2
= 4𝑥( 3𝑥 + )
12𝑥 2
Step 3: The HCF multiplied by the second term in the bracket needs
to give us 8𝑥
= 4𝑥( 3𝑥 + 2 )
8𝑥
So, we have our answer of 4𝑥(3𝑥 + 2)
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Exercise A5
1.
24𝑥 + 36
2.
56𝑦 − 21
3.
20𝑓 − 25
4.
10𝑥 + 45
5.
16𝑤 2 + 20𝑤
6.
25ℎ2 + 20ℎ3
7.
16𝑤 2 + 20𝑤
8.
18𝑦 5 − 27𝑦 2
9.
35𝑒 12 + 10𝑒 6
10.
9𝑡 13 − 24𝑡 7
You can check you have not made a mistake by expanding your
answer.
You can find more questions in the Exam Questions section.
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Factorising to a double bracket
These questions will be clear as you will be given three terms. The
first will contain a 'letter squared', the second will contain a 'letter'
and the third will be a number. Here's an example.
Example 1
𝑥 2 + 5𝑥 + 6
To factorise this, we need to find the same two numbers that;
1. Multiply to give the last term (6)
2. Add to give the number in the middle term (5)
Then place those numbers in the following:
(𝑥 ____)(𝑥 ____)
In this example, the two numbers are +3 and +2 (3x2=6 and 3+2=5)
So,
(𝑥 + 3)(𝑥 + 2)
Example 2
Factorise
𝑥 2 + 11𝑥 + 24
Listing all the factor pairs of the last number gives all possible
answers. (E.g. 24x1,12x2, 3x8, 6x4)
The numbers needed here are 3 and 8 (8x3=24 and 8+3=11)
(𝑥 + 8)(𝑥 + 3)
You can check your answer by Expanding and Simplifying.
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Consider this more difficult example
Example 3
Factorise:
𝑥 2 + 2𝑥 − 15
Listing all factor pairs of the final number is the place to start. (15x1,
5x3). 5 and 3 could work if we use a minus sign on one of the
numbers. (5 -3 = 2 and 5 x -3 = -15)
(𝑥 + 5)(𝑥 − 3)
Here's one final example to consider:
Example 4
Factorise:
𝑥 2 − 10𝑥 + 24
Again, begin by listing all factor pairs (24x1, 12x2, 8x3, 6x4).
The only way to get -10 is to use the 6 and 4 but make both
numbers negative. (-6 - 4 = -10 and -6 x -4 =24).
(𝑥 − 6)(𝑥 − 4)
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Exercise A6
1.
𝑥 2 + 12𝑥 + 11
2.
𝑥 2 + 11𝑥 + 30
3.
𝑥 2 + 16𝑥 + 60
4.
𝑥 2 − 8𝑥 + 12
5.
𝑥 2 − 15𝑥 + 56
6.
𝑥 2 − 4𝑥 − 12
7.
𝑥 2 + 5𝑥 − 14
Exam Questions A2
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