We've already looked at how to add and subtract algebraic fractions, and know that it involves making sure both fractions have the same denominator. To convert fractions so that they have a common denominator, we need to look at their factors, which can allow us to find the lowest common multiple (LCM) of the original denominators. That's why factorisation becomes very useful when we have complicated algebraic expressions as our denominators.

Some Quick Revision

Let's have a look at the example $\frac{8}{x^2}+\frac{x+2}{2x}$8x2+x+22x. Right now the two denominators are not the same, with the factors of the first being $x$x and $x$x, while the second being $2$2 and $x$x.

We can straight away see that the LCM of the two is $2x^2$2x2, which will be our new denominator. That means to convert the two fractions we would need to multiply the first by $\frac{2}{2}$22 and the second by $\frac{x}{x}$xx.

So our resulting problem is $\frac{8\times2}{2x^2}+\frac{\left(x+2\right)x}{2x^2}=\frac{16+\left(x+2\right)x}{2x^2}$8×22x2+(x+2)x2x2=16+(x+2)x2x2.

Expanding we then get $\frac{16+x^2+2x}{2x^2}$16+x2+2x2x2 which is our final answer.

Using Factorisation

Let's have a look at some examples to see how factorisation can help us. We can use the various factorisation techniques we have learnt previously.

Examples

Question 1

Simplify the following expression, giving your answer in fully factorised form:

$\frac{x}{x^2-16}-\frac{12}{x+4}$xx2−16−12x+4

Question 2

Simplify the following expression:

$\frac{x+8}{x^2+19x+88}+\frac{x-7}{\left(x+11\right)\left(x+7\right)}$x+8x2+19x+88+x−7(x+11)(x+7)

Question 3

Factorise the denominators and then simplify:

$\frac{2}{x^2+11x+18}-\frac{5}{x^2-4}$2x2+11x+18−5x2−4

Factorising algebraic fractions (add/sub)