Algebraic fractions

An algebraic fraction is a fraction that has a polynomial in both the numerator and the denominator. We usually try to simplify an algebraic fraction into a shorter, neater expression.

The basics

So once more, an algebraic fraction has this form:

$$\frac{\text{polynomial}}{\text{polynomial}}$$

An example of an algebraic fraction is the following expression:

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

The task is then to simplify this expression. To simplify it we use the same methods as when we simplify polynomials, so we use, for example, taking out a common factor and various useful formulae. The next section summarises the most common steps.

The denominator of an algebraic fraction should contain a variable. For example, we do not call the following expression an algebraic fraction:

$$\frac{2x+3}{2}$$

For an algebraic fraction we also usually state the restrictions, that is, the values for which the fraction makes sense. The denominator cannot be equal to zero, because, as we know, we cannot divide by zero.

Techniques

When simplifying expressions we use many techniques, and I will try to put together a summary of the common ones. First of all there is cancelling. If the numerator and denominator are both products of factors, we can cancel:

$$\frac{2a}{5a}=\frac{2}{5}$$

Here we cancelled the variable a. If the fraction looks like this

$$\frac{2a+10}{5a}\ne \frac{2+10}{5},$$

we cannot cancel. The numerator is a sum, so cancelling is not allowed. It often happens that the numerator or denominator is a sum, but we can take out a common factor and then cancel. For example, in the fraction

$$\frac{5a+6ab}{2a}$$

the numerator is a sum, but we can take out a and then cancel:

$$\frac{5a+6ab}{2a}=\frac{a(5+6b)}{2a}=\frac{5+6b}{2}.$$

Another common step is to factorise something using a formula. Above we had the algebraic fraction

$$\frac{x^2-1}{x+1}.$$

If we apply the formula

$$a^2-b^2=(a+b)(a-b),$$

to the numerator, we get a new algebraic fraction

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

and here we can cancel the whole bracket (x + 1):

$$\frac{(x+1)(x-1)}{(x+1)}=\frac{x-1}{1}=x-1.$$

Then, of course, there are all the usual steps: adding and multiplying polynomials, using the laws of indices, adding fractions and so on.

First example

The question:

$$\frac{6a+12b}{2a+4b}$$

We can see that we can take out a 2 from both the numerator and the denominator, and then cancel it:

$$\frac{6a+12b}{2a+4b}=\frac{2(3a+6b)}{2(a+2b)}=\frac{3a+6b}{a+2b}$$

In the numerator we can also take out a 3:

$$\frac{3a+6b}{a+2b}=\frac{3(a+2b)}{a+2b}$$

We can cancel the whole expression a + 2b, leaving just the 3:

$$\frac{3(a+2b)}{a+2b}=\frac{3}{1}=3$$

The restrictions for this algebraic fraction:

$$a+2b\ne 0$$

Second example

The question:

$$(\frac{x+1}{x+2}-\frac{x-1}{x-2})\cdot \frac{x^2-4}{2x}$$

In the first step we subtract the fractions in the inner bracket:

$$\frac{(x+1)(x-2)-(x-1)(x+2)}{(x+2)(x-2)}\cdot \frac{x^2-4}{2x}$$

Next we multiply out the brackets in the denominator. Sharper students may notice that we can apply the formula a² − b²:

$$\frac{(x+1)(x-2)-(x-1)(x+2)}{x^2-4}\cdot \frac{x^2-4}{2x}$$

The denominator of the first fraction is the same expression as the numerator of the second fraction, and since we are multiplying these fractions, we can cancel this expression:

$$\frac{(x+1)(x-2)-(x-1)(x+2)}{1}\cdot \frac{1}{2x}=\frac{(x+1)(x-2)-(x-1)(x+2)}{2x}$$

Now we have to do some hard graft: expand the brackets in the numerator, collect like terms and finally cancel.

$$\frac{x^2-2x+x-2-(x^2+2x-x-2)}{2x}=\frac{x^2-2x+x-2-x^2-2x+x+2}{2x}=$$

$$=\frac{-2x}{2x}=\frac{-1}{1}=-1$$

We find the restrictions from the question. There are three fractions in it, and none of the denominators can be equal to zero.

$$\begin{array}{rcl}x& \ne & 2\\ x& \ne & -2\\ x& \ne & 0\end{array}$$

Third example

The next algebraic fraction:

$$(\frac{a+b}{ab}-\frac{a-b}{ab}+2)\cdot (\frac{a}{a+1}-\frac{a}{b+1})$$

Once again we start with the tedious job of adding and subtracting the expressions in the brackets. The first difference is easy, because both fractions have the same denominator.

$$(\frac{(a+b)-(a-b)}{ab}+2)\cdot \left(\frac{a(b+1)-(a+1)a}{(a+1)(b+1)}\right)$$

We expand the brackets and collect like terms in the numerators:

$$(\frac{2b}{ab}+2)\cdot \left(\frac{ab+a-a^2-a}{(a+1)(b+1)}\right)$$

In the first fraction we cancel the variable b, and we write 2 as a fraction over a so that we can add it to the first fraction:

$$(\frac{2}{a}+\frac{2a}{a})\cdot \frac{ab-a^2}{(a+1)(b+1)}$$

We add the fractions in the first bracket and take out a in the numerator of the second fraction.

$$\frac{2+2a}{a}\cdot \frac{a(b-a)}{(a+1)(b+1)}$$

Now we can cancel a:

$$\frac{2+2a}{\boxed{\text{a}}}\cdot \frac{\boxed{\text{a}}(b-a)}{(a+1)(b+1)}=(2+2a)\frac{b-a}{(a+1)(b+1)}$$

Now we just multiply the expression on the left by the fraction.

$$\frac{(2+2a)(b-a)}{(a+1)(b+1)}$$

From the first bracket we take out a 2:

$$\frac{2(1+a)(b-a)}{(a+1)(b+1)}$$

Now we can cancel (1 + a) (the denominator has the same expression, just written the other way round as a + 1).

$$\frac{2(b-a)}{b+1}$$

This is the final simplified algebraic fraction, and there is nothing sensible left to do with it. The restrictions:

$$\begin{array}{rcl}ab& \ne & 0\to a,b\ne 0\\ a& \ne & -1\\ b& \ne & -1\end{array}$$

Algebraic division

You can also simplify algebraic fractions using the algorithm for dividing a polynomial by a polynomial. This can be useful when none of the usual methods for simplifying expressions can be used.