# word ```Academic Skills Advice
Factorisation &amp; Function Notation
Factorisation:
Some expressions can be simplified by factorising. This means taking out (dividing by)
any common factors (i.e. numbers or letters that are in every term of the expression) and
putting everything that’s left into brackets. This is the opposite to “multiplying out brackets”
covered in “Basic Algebra”.
A reminder of cancelling fractions:
To be able to divide each term in the expression by the common factor you need to
remember how to divide fractions (covered in lesson 1). Here are a few examples:
3𝑏
3𝑏
=1
4𝑥𝑦
8𝑥
=
10𝑎2 𝑏 2
5𝑎𝑏 5
(top &amp; bottom are the same so they cancel leaving 1)
𝑦
(4 goes into 8 twice, the ‘𝑥’ at the top and bottom cancel)
2
=
2𝑎
𝑏3
(10&divide;5=2 and the rules of indices say that 𝑎2 &divide; 𝑎 = 𝑎2−1 = 𝑎
and 𝑏 2 &divide; 𝑏 5 = 𝑏 2−5 = 𝑏 −3 (=
You could also do this one by cancelling top &amp; bottom:
10𝑎2 𝑏 2
5𝑎𝑏 5
1
))
𝑏3
2
10&times;𝑎&times;𝑎&times;𝑏&times;𝑏
= 5&times;𝑎&times;𝑏&times;𝑏&times;𝑏&times;𝑏&times;𝑏 =
2𝑎
𝑏3
Examples of Factorising:

Simplify the following:
𝟔𝒂 + 𝟔𝒃
Both terms will divide by 6 so put 6 outside the bracket and work out what will stay
inside the bracket (by dividing everything by 6).

Starting expression:
6𝑎 + 6𝑏
The 6’s will cancel top &amp; bottom
6( 6 + 6 )
So we end up with:
6(𝑎 + 𝑏)
Simplify the following:
6𝑎
6𝑏
You could check this is correct
by multiplying out the bracket
and making sure you get back to
where you started.
𝟕𝒙 + 𝟐𝟏𝒙𝒚
Both terms will divide by 7 and 𝑥 so put 7𝑥 outside the bracket and work out what will
stay inside the bracket (by dividing everything by 7𝑥).
Starting expression:
7𝑥 + 21𝑥𝑦
The 7𝑥 will cancel top &amp; bottom
7𝑥 (7𝑥 + 7𝑥 )
So we end up with:
7𝑥(1 + 3𝑦)
&copy; H Jackson 2011 / ACADEMIC SKILLS
7𝑥
21𝑥𝑦
Multiply out the bracket to
check it’s correct.
1

Simplify the following:
Collect like terms:
So we have:
𝟖𝒙 + 𝟑𝒙𝒚 + 𝟗𝒙𝒚 − 𝟒𝒙
8𝑥 − 4𝑥 = 4𝑥
3𝑥𝑦 + 9𝑥𝑦 = 12𝑥𝑦
4𝑥 + 12𝑥𝑦
Now we can factorise because each term will divide by 4 and 𝑥. So we divide every term
by 4𝑥 and put it outside the bracket.

Starting expression:
4𝑥 + 12𝑥𝑦
The 4𝑥 will cancel top &amp; bottom
4𝑥 (4𝑥 + 4𝑥 )
So we end up with:
4𝑥(1 + 3𝑦)
4𝑥
Multiply out the bracket to
check it’s correct.
12𝑥𝑦
Simplify the following expression: 𝟑𝒂𝒃𝟐 + 𝟒𝒂𝟐 𝒃 – 𝒃𝟐 𝒂 – 𝟓𝒂𝒃 – 𝟔𝒂𝟐 𝒃
This looks more complicated but remember you can collect any terms that are the same.
Notice that 𝑎𝑏 2 is the same as 𝑏 2 𝑎 (just written the other way round) but is not the same
as 𝑎2 𝑏.
Collect like terms:
3𝑎𝑏 2 – 𝑏 2 𝑎 = 2𝑎𝑏 2
4𝑎2 𝑏 – 6𝑎2 𝑏 = −2𝑎2 𝑏
−5𝑎𝑏 (there is no other 𝑎𝑏 term)
So we have:
2𝑎𝑏 2 − 2𝑎2 𝑏 − 5𝑎𝑏
Now we have an expression that we can factorise as each term includes an ‘𝑎’ and a ‘𝑏’ so
we can divide each term by 𝑎𝑏 and put it outside the bracket.
Starting expression:
2𝑎𝑏 2 − 2𝑎2 𝑏 − 5𝑎𝑏
The 𝑎𝑏 will cancel top &amp; bottom
𝑎𝑏 ( 𝑎𝑏 − 𝑎𝑏 − 𝑎𝑏 )
So we end up with:

2𝑎𝑏
2
2𝑎2 𝑏
5𝑎𝑏
Check it’s correct.
𝑎𝑏(2𝑏 − 2𝑎 − 5)
Simplify the following expression: 𝒙𝟑 𝒚𝟐 + 𝒙𝟐 𝒚𝟑 + 𝒙𝟐 𝒚
Now we need to decide on the highest powers we can divide by. Every term will divide by
an ‘𝑥 2 ’ and a ‘𝑦’ so we can divide each term by 𝑥 2 𝑦 and put it outside the bracket. N.b. we
cannot divide by 𝑦 2 as the last term has a lower power in (𝑦).
Starting expression:
𝑥3𝑦2 + 𝑥2𝑦3 + 𝑥2𝑦
1
2
2
1
𝑥3𝑦2
2
𝑥2𝑦3
The 𝑥 𝑦 will cancel top &amp; bottom
𝑥 𝑦 ( 𝑥2𝑦 +
So we end up with:
𝑥 2 𝑦 (𝑥𝑦 + 𝑦 2 + 1)
&copy; H Jackson 2011 / ACADEMIC SKILLS
𝑥2𝑦
+
𝑥2𝑦
𝑥2𝑦
)
Check it’s correct.
2
Function Notation:
Function notation is another way of writing equations. Instead of writing “𝑦 =”, we write
“𝑓(𝑥) =” or “𝑔(𝑥) =” etc. We say “𝑓 of 𝑥” or “𝑔 of 𝑥”.
You might see a function written as:
𝒇(𝒙) = 𝟑𝒙 − 𝟒
OR
𝒇: 𝒙 → 𝟑𝒙 − 𝟒
𝑓(𝑥) and 𝑓: 𝑥 both just mean “a function of 𝒙” and the right hand side is the algebra
explaining what to do. i.e. do 3 &times; 𝑥 then take away 4. (Remember 𝑓(𝑥) does not mean 𝑓 &times;
(𝑥).)
Evaluating a Function:
When asked to evaluate a function we just replace the 𝑥 with whatever we are finding the
function of. It’s always a good idea to use brackets so that you are careful with any signs.
Examples:

Given the function: 𝒇(𝒙) = 𝟓𝒙 + 𝟐
Find 𝒇(𝟔)
(we say “𝒇 of 6”)
This just means that wherever you see an 𝑥 you replace it with 6.
𝑓(𝑥) = 5𝑥 + 2
𝑓(6) = 5(6) + 2
𝑓(6) = 5 &times; 6 + 2
𝑓(6) = 32
We have used 𝑓 to denote the function but we can also use other letters to mean a
different function.

Given the function: 𝒈(𝒙) = 𝟒𝒙𝟐 − 𝟖𝒙
Find 𝒈(−𝟑)
(we say “𝒈 of – 𝟑”)
Now wherever you see an 𝑥 replace it with −3.
𝑔(𝑥) = 4𝑥 2 − 8𝑥
𝑔(−3) = 4(−3)2 − 8(−3)
𝑔(−3) = 4 &times; (−3)2 − 8 &times; (−3)
𝑔(−3) = 60

Given the function: 𝒉(𝒙) = 𝟕𝒙 − 𝟓
Find 𝒉(𝟏𝟎)
(we say “𝒉 of 10”)
Wherever you see an 𝑥 replace it with 10.
ℎ(𝑥) = 7𝑥 − 5
ℎ(10) = 7(10) − 5
ℎ(10) = 70 − 5
ℎ(10) = 65