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### Algebraic fractions

An algebraic fraction is a fraction containing pronumerals. Later, when functions are introduced, an algebraic fraction in which both the numerator and the denominator are polynomials will be referred to as a rational function.

Algebraic fractions are manipulated and simplified using exactly the same rules that apply to ordinary fractions. In particular, to add or subtract them we need to find (lowest) common denominators. In many instances factoring needs to be done first so that these common factors become apparent.

For example, to simplify

$\frac{4}{x^2+x} - \frac{3}{x^2-1}$

we need to factor each denominator using the techniques outlined above. Thus we write

$\frac{4}{x^2+x} - \frac{3}{x^2-1} = \frac{4}{x(x+1)}- \frac{3}{(x-1)(x+1)}.$

It is now apparent that the lowest common denominator is $$x(x+1)(x-1)$$. Hence,

\begin{align*} \frac{4}{x(x+1)}- \frac{3}{(x-1)(x+1)} &= \frac{4(x-1)-3x}{x(x+1)(x-1)}\\ &= \frac{x-4}{x(x^2-1)}. \end{align*}

It is worth mentioning in passing that we assume in all these steps that $$x$$ does not take any of the values 0, 1 or $$-1$$, since that would make the denominators equal to zero. While being aware of this, we generally do not need to make a big issue of it. It is tacitly assumed in all our calculations.

Factoring may also be very useful when multiplying two algebraic fractions, since we can then see any cancellation that may occur.

For example, to multiply

$\frac{3x}{x^2-9} \times \frac{x+3}{3x^2-6x}$

we factor, where possible, first:

$\frac{3x}{x^2-9} \times \frac{x+3}{3x^2-6x} = \frac{3x}{(x-3)(x+3)} \times \frac{(x+3)}{3x(x-2)}.$

Cancelling out the factors $$3x$$ and $$x+3$$, we have

$\frac{3x}{(x-3)(x+3)} \times \frac{(x+3)}{3x(x-2)} = \frac{1}{(x-3)(x-2)}.$
##### Exercise 6

Simplify

$\frac{8x}{x^2+5x+6} - \frac{5x}{x^2+3x+2} -\frac{3x}{x^2+4x+3}.$

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