Note that the reason for \(x\) not being allowed to be 0, 2 or -1 is so that none of the factors of the denominator can possibly be zero. Division by zero is impossible.

In this example, both the numerator and denominator are already fully factorised. So we do not have to do any more factorising ourselves.

There are common factors in both the numerator and denominator of \(x\) and \(\left(x+1\right)\). These can cancel.

So we get the following working:

$$ \begin{eqnarray} \small\frac{{x^2}(x+1)}{x(x-2)(x+1)}\ &=&\small\ \frac{{x^\cancel{2}}\cancel{(x+1)}}{\cancel{x}(x-2)\cancel{(x+1)}}\\[8pt] &=&\small\ \frac{x}{x-2} \end{eqnarray} $$

In this example, we need to factorise both the numerator and denominator to see if there are any common factors that we can cancel.

$$ \begin{eqnarray} \small\frac{x^2+3x-4}{x^2-x}\ &=&\small\ \frac{(x-1)(x+4)}{x(x-1)}\\[8pt] &=&\small\ \frac{\cancel{(x-1)}(x+4)}{x\cancel{(x-1)}}\\[8pt] &=&\small\ \frac{x+4}{x} \end{eqnarray} $$

Just like adding numerical fractions, we need a common denominator.

The lowest common multiple of 4 and 6 is 12.

$$ \begin{eqnarray} \small\frac{3a}{4}+\frac{5a}{6}\ &=&\small\ \frac{9a}{12}+\frac{10a}{12}\\[8pt] &=&\small\ \frac{19a}{12} \end{eqnarray} $$

Just like subtracting numerical fractions, we need a common denominator.

With numerical fractions, we would use the lowest common multiple of the denominators. But in this example, we don't know the value of \(x\), so we don't know if it's a multiple of 5.

Because of this, we just have to multiply the denominators and use \(5x\) as the common denominator.

$$ \begin{eqnarray} \small\frac{3}{x}-\frac{2x}{5}\ &=&\small\ \frac{15}{5x}-\frac{2x^2}{5x}\\[8pt] &=&\small\ \frac{15-2x^2}{5x} \end{eqnarray} $$

Just like multiplying numerical fractions, we first try to cancel any common factors.

How do we find common factors? By factorising, of course!

$$ \begin{eqnarray} &\ &\small (x+3) \times \frac{x^2+7}{x^2-9} \\[12pt] &=&\small\ \frac{x+3}{1} \times \frac{x^2+7}{(x-3)(x+3)} \\[12pt] &=&\small\ \frac{\cancel{x+3}}{1} \times \frac{x^2+7}{(x-3)(\cancel{x+3})} \\[12pt] &=&\small\ \frac{x^2+7}{x-3} \end{eqnarray} $$

Just like dividing numerical fractions, we multiply by the reciprocal.

$$ \begin{eqnarray} \small\frac{p}{2t^2} \div \frac{p^2}{4t}\ &=&\ \small\frac{p}{2t^2}\times \frac{4t}{p^2}\\[10pt] &=&\small\ \frac{\cancel{p}}{\cancelto{1}{2}t^\cancel{2}}\times \frac{\cancelto{2}{4}\cancel{t}}{p^\cancel{2}}\\[10pt] &=&\small\ \frac{2}{pt} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{7}{x+5}-\frac{3}{x}\ &=&\small\ \frac{7x}{x(x+5)}-\frac{3(x+5)}{x(x+5)}\\[10pt] &=&\small\ \frac{7x-3(x+5)}{x(x+5)}\\[10pt] &=&\small\ \frac{7x-3x-15}{x(x+5)}\\[10pt] &=&\small\ \frac{4x-15}{x(x+5)} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{5t}{s} \div \frac{t}{2s^2}\ &=&\small\ \frac{5t}{s}\times \frac{2s^2}{t}\\[10pt] &=&\small\ \frac{5\cancel{t}}{\cancel{s}}\times \frac{2s\cancel{^2}}{\cancel{t}}\\[10pt] &=&\ 10s \end{eqnarray} $$

$$ \begin{eqnarray} &\ &\small\frac{3}{x-2}+\frac{5}{x+1}\ \\[10pt] &=&\small\ \frac{3(x+1)}{(x-2)(x+1)}+\frac{5(x-2)}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{3(x+1)+5(x-2)}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{3x+3+5x-10}{(x+1)(x-2)}\\[10pt] &=&\small\ \frac{8x-7}{(x+1)(x-2)} \end{eqnarray} $$

$$ \begin{eqnarray} &\ &\small\frac{4}{x+2}-\frac{3}{x-4}\ \\[10pt] &=&\small\ \frac{4(x-4)}{(x-4)(x+2)}-\frac{3(x+2)}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{4(x-4)-3(x+2)}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{4x-16-3x-6}{(x-4)(x+2)}\\[10pt] &=&\small\ \frac{x-22}{(x-4)(x+2)} \end{eqnarray} $$

This numerator has a common factor of \(x.\) The denominator is a trinomial. So we factorise as follows:

$$ \begin{eqnarray} \small\frac{x^2-4x}{x^2+x-20}&=&\small\frac{x(x-4)}{(x-4)(x+5)}\\[10pt] &=&\small\ \frac{x}{x+5} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{3}{a^2}-\frac{2}{a}\ &=&\small\ \frac{3}{a^2}-\frac{2a}{a^2}\\[10pt] &=&\small\ \frac{3-2a}{a^2} \end{eqnarray} $$

$$ \begin{eqnarray} 4x^2-25 &=&(2x)^2-5^2\\[10pt] &=& (2x-5)(2x+5) \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{4x^2-25}{2x^2-x-10} &=&\small\ \frac{(2x-5)(2x+5)}{(x+2)(2x-5)}\\[10pt] &=&\small\ \frac{2x+5}{x+2} \end{eqnarray} $$

$$ \begin{eqnarray} &\ &\small\frac{n}{n^2-4} \div \frac{3}{n-2}\\[10pt] &=&\small\ \frac{n}{(n-2)(n+2)}\times \frac{n-2}{3}\\[10pt] &=&\small\ \frac{n}{\cancel{(n-2)}(n+2)}\times \frac{\cancel{n-2}}{3}\\[10pt] &=&\small\ \frac{n}{3(n+2)} \end{eqnarray} $$

$$ \begin{eqnarray} &\ &\small\frac{4}{x-2}-\frac{3}{x+5}\ \\[10pt] &=&\small\ \frac{4(x+5)}{(x-2)(x+5)}-\frac{3(x-2)}{(x+5)(x-2)}\\[10pt] &=&\small\ \frac{4(x+5)-3(x-2)}{(x-2)(x+5)}\\[10pt] &=&\small\ \frac{4x+20-3x+6}{(x-2)(x+5)}\\[10pt] &=&\small\ \frac{x+26}{(x-2)(x+5)} \end{eqnarray} $$

$$ \begin{eqnarray} \small\frac{7}{x-3}-\frac{2}{x}\ &=&\small\ \frac{7x}{x(x-3)}-\frac{2(x-3)}{x(x-3)}\\[10pt] &=&\small\ \frac{7x-2(x-3)}{x(x-3)}\\[10pt] &=&\small\ \frac{7x-2x+6}{x(x-3)}\\[10pt] &=&\small\ \frac{5x+6}{x(x-3)} \end{eqnarray} $$

This numerator is a difference of two squares. The denominator is a trinomial. So we factorise as follows:

$$ \begin{eqnarray} \small\frac{x^2-16}{x^2+x-20}&=&\small\frac{(x+4)(x-4)}{(x-4)(x+5)}\\[10pt] &=&\small\ \frac{x+4}{x+5} \end{eqnarray} $$