Broken expressions

A fraction is a fraction that has a polynomial in both the numerator and denominator. We typically try to simplify a fractional expression into some shorter, nicer expression.

Basics

So once again, a fractional expression has the following form:

$$\frac{\mbox{ multi-part }}{\mbox{ multi-part }}$$

So an example of a refracted expression might be the following expression:

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

The challenge then is to simplify this expression. When simplifying, we use the same techniques we use when modifying polynomials, so we'll use things like howling and various useful formulas. The next section summarizes the most commonly used modifications.

A fractional expression should contain some variable in the denominator; for example, we don't say this is a fractional expression:

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

For a refracted expression, we also usually specify the conditions under which the refracted expression makes sense. For a refracted expression, the denominator must not be equal to zero because, as we know, it is not divisible by zero.

Techniques

There are many techniques we use when modifying expressions, so I'll try to write down a summary of some common ones. First of all, there is the truncation of fractions. So if we have expressions in both the numerator and denominator that we are just multiplying between, we can shorten:

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

Here we have truncated the variable a. If the fraction looked like this

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

then we can't truncate. The numerator is in the sum, so truncation is not allowed. It often happens that even though the denominator or numerator of a fraction is in the sum form, we can make something in the expression and then we can shorten. For example, in a fraction

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

we have the numerator in the form of a sum, but we can print a and then truncate

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

A common modification is also to decompose something according to some formula. Above we had a fractional expression

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

If we apply the formula to the numerator

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

we get a new fractional expression

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

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

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

Then of course various addition and multiplication of polynomials, adjusting powers, adding fractions, etc.

The first example

Example assignment:

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

We can see that we can set off a two in the numerator and denominator, which we can then truncate:

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

In the numerator, we can also print a three:

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

We can truncate the whole expression a + 2b and we are left with a three:

$$\frac{3(a+2b)}{a+2b}=\frac31=3$$

The terms of this broken expression:

$$a+2b\ne0$$

Second example

Example Assignment:

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

In the first step, subtract the fractions in the inner parenthesis:

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

Next, we multiply the parenthesis in the denominator. Astute students may notice that we can apply the formula a² − b²:

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

We have the same expression in the denominator of the first fraction as in the numerator of the second fraction, and since we are multiplying these fractions, we can use this expression to multiply:

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

Now we need to apply a little rough work and multiply and add the parentheses in the numerator and finally simply multiply:

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

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

We will determine the terms according to the assignment. There are a total of three fractions, none of the denominators can equal zero.

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

The third example

Entering another fractional expression:

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

So first, we again boringly add and subtract the expressions in parentheses. The first difference will be easy because both fractions have the same denominator.

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

We multiply and add the expressions in the numerators:

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

In the first fraction, we truncate the variable b and extend the two by a, so that we can add it to the first fraction:

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

We add the fractions in the first parenthesis and in the second fraction we set off a in the numerator.

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

So, now we can truncate a:

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

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

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

From the first parenthesis, we'll pick out a two:

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

Now we can truncate (1 + a) (the same expression in the denominator, just in reverse a + 1).

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

And that's the final simplified fractional expression, there's nothing sensible to do with this expression. Conditions:

$$\begin{eqnarray} ab&\ne&0\rightarrow a,b\ne0\\ a&\ne&-1\\ b&\ne&-1 \end{eqnarray}$$

Division of polynomials

You can also solve the broken expressions using the algorithm for dividing polynomials by polynomials. This can be convenient when you cannot use some classical means of modifying expressions.