Algebraic Fractions

Algebra and Sequences

An algebraic fraction has an expression on the top, the bottom, or both
Every rule that works for ordinary fractions works here, because the letters simply stand for numbers
Simplifying means cancelling, and cancelling is only ever allowed between factors
So the first move is always to factorise the top and the bottom as far as they will go
A factor that appears above and below then divides out, whether it is a single term or a whole bracket
Marks are given for each correct factorisation, so a part-finished attempt still scores

A term that is added on is not a factor, so it cannot be cancelled
- In (4x + 8) ÷ (4x − 1) the 4x cannot be removed, because the top is 4(x + 2) and the bottom has no factor of 4
Test any cancelling by asking whether the thing being removed multiplies everything above and everything below
Where neither part factorises and no bracket is shared, there is nothing to cancel and the fraction stands as it is
When one of the two expressions is hard to factorise, factorise the easier one first: the same bracket almost always turns up in the other

Worked example

Factorising both parts before cancelling

Simplify (2x² − 14x) ÷ (x² − 9x + 14).

Solution:

Nothing can be cancelled while the top and bottom are sums, so factorise both first
The top has a common factor in each term: 2x² and 14x both contain 2x
Top = 2x(x − 7)
The bottom is a quadratic with no number in front of x², so find two numbers multiplying to +14 and adding to −9
Both must be negative to give a positive product and a negative sum: −7 and −2
Bottom = (x − 7)(x − 2)
The fraction now reads 2x(x − 7) ÷ [(x − 7)(x − 2)]
(x − 7) is a factor of both the top and the bottom, so it divides out
The answer is 2x ÷ (x − 2)
Only whole brackets cancel: the x in (x − 2) cannot be cancelled against the 2x on top, because it is a term rather than a factor
Each correct factorisation scores its own mark, so both lines are worth writing even if the cancelling is missed