Simplifying algebraic fractions

In a nutshell

Algebraic fractions are fractions that can be represented as the ratio of two polynomials. Prior to operating with them, it is useful to factorise the numerator and denominator in order to simplify the process. 

Cancelling factors in algebraic fractions

Procedure

Example 1

Simplify this fraction as far as possible $$\cfrac{x^2 +9 x + 20}{x+4}$$​

First you factorise numerator and denominator. The denominator is already factorised, so you only need to focus on the numerator. 

The numerator $$x^2 + 9x +20$$ factorises into $$(x+4)(x+5)$$. Therefore, you have 

$$\cfrac{x^2 +9 x + 20}{x+4} = \cfrac{(x+4)(x+5)}{x+4}$$​​

Now cancel the common factors. 

​$$\cfrac{(x+4)(x+5)}{x+4} = \cfrac{\cancel{(x+4)}(x+5)}{\cancel{x+4}}$$​​

In conclusion, $$\underline{\cfrac{x^2 +9 x + 20}{x+4} = x+5}$$​​

Example 2

Simplify this fraction $$\cfrac{8x^3 - 4x^2 -6x}{2x}$$​

You have that $$x$$ is a common factor, so it can be eliminated. This is also true of $$2$$:

$$\cfrac{8x^3 - 4x^2 -6x}{2x} = \cfrac{\cancel x (8x^2 - 4x- 6)}{2\cancel x}= \cfrac{\cancel 2 (4x^2 -2x -3)}{\cancel 2} = 4x^2 -2x -3$$​​

As $$4x^2 -2x -3$$ cannot be factorised into an expression $$(Ax+B)(Cx+D)$$ where $$A$$, $$B$$, $$C$$ and $$D$$ are integers, this is the fully factorised expression.

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Therefore, $$\underline{\cfrac{8x^3 - 4x^2 -6x}{2x} = 4x^2 -2x -3}$$