Rational Expressions

A rational expression is something of the form

$$\dfrac{\hbox{polynomial}}{\hbox{polynomial}}.$$

Rational expressions can often be simplified by factoring the top and bottom, then cancelling any common factors. However, cancelling common factors can change the domain of the expression, and in some cases, this is important. In many of the examples below, I'll just do the cancellation without comment --- but first, here are some examples which show what the issue is.


Example. The following cancellation is valid for $x \ne 1$ :

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

If $x = 1$ , then the common factor $x - 1$ is equal to 0, so in that case it is like simplifying $\dfrac{0}{0}$ to get 1 (which isn't right).

Notice that the original expression $\dfrac{x^2 - 1}{x - 1}$ is undefined if $x = 1$ (since $x =
   1$ makes the bottom 0), but the final expression $x + 1$ is defined when $x = 1$ (it's equal to 2).


Example. The following cancellation is valid for $x \ne -2$ :

$$\dfrac{x^2 + 4x + 4}{x^2 + 5x + 6} = \dfrac{(x + 2)^2}{x + 2)(x + 3)} = \dfrac{x + 2}{x + 3}.$$

When $x = -2$ , the common factor $x
   + 2$ is equal to 0.


Example. The following cancellation is valid for $x \ne 1$ :

$$\dfrac{x^2 - 7x - 8}{x^2 + 3x + 2} = \dfrac{(x - 8)(x + 1)}{(x + 2)(x + 1)} = \dfrac{x - 8}{x + 2}.$$

When $x = -1$ , the common factor $x
   + 1$ is equal to 0.


Example. The following cancellation is valid for $x \ne 0$ :

$$\dfrac{x^4 - x^2}{x^3 + 5x} = \dfrac{x^2(x^2 - 1)}{x^2(x + 5)} = \dfrac{x^2 - 1}{x^2 + 5}.$$

When $x = 0$ , the common factor $x^2$ is equal to 0.


In working with fractions --- in fact, in doing algebra in general --- it is easy to do things which aren't allowed. What operations are "allowed"? Generally, an operation is allowed if the resulting equality is an identity: An equation which is true for all values of the variables where both sides of the equation are defined.

If you're not sure whether an operation is valid, the best thing is to try to prove it using valid algebra rules you already know. But it's easy to show that an operation is invalid: Just find a value of the variable for which the two sides aren't equal. This shows that the equation is not an identity.


Example. Prove that the following is not an identity:

$$\dfrac{2}{x + 2} \buildrel {\rm ?} \over = \dfrac{2}{x} + \dfrac{2}{2}.$$

If $x = 2$ , then

$$\dfrac{2}{x + 2} = \dfrac{1}{2}, \quad\hbox{but}\quad \dfrac{2}{x} + \dfrac{2}{2} = 2.\quad\halmos$$


Example. Prove that the following is not an identity:

$$\dfrac{x + 2}{x + 4} \buildrel {\rm ?} \over = \dfrac{x}{x} + \dfrac{2}{4}.$$

If $x = 2$ , then

$$\dfrac{x + 2}{x + 4}= \dfrac{4}{6}, \quad\hbox{but}\quad \dfrac{x}{x} + \dfrac{2}{4} = \dfrac{3}{2}.\quad\halmos$$


You can often simplify a complex fraction by multiplying the top and bottom by the same thing. Usually, you do this to clear denominators if you have smaller fractions on the top or bottom of a larger fraction.


Example. Simplify $\dfrac{\dfrac{1}{x} - 1}{x - 1}$ .

In this case, I multiply the top and bottom by x to clear the $\dfrac{1}{x}$ on the top:

$$\dfrac{\dfrac{1}{x} - 1}{x - 1} = \dfrac{\dfrac{1}{x} - 1}{x - 1}\cdot \dfrac{x}{x} = \dfrac{1 - x}{x(x - 1)} = \dfrac{-(x - 1)}{x(x - 1)} = -\dfrac{1}{x}.$$

Multiplying by $\dfrac{x}{x}$ is valid, because $\dfrac{x}{x} = 1$ (as long as $x \ne 0$ ), and multiplying by 1 doesn't change the value.

You can also do this problem by combining $\dfrac{1}{x}$ and 1 over a common denominator.


Example. Simplify $\dfrac{\dfrac{1}{3} - \dfrac{1}{x + 5}}{x^2 - 4}$ .

$$\dfrac{\dfrac{1}{3} - \dfrac{1}{x + 5}}{x^2 - 4} = \dfrac{\dfrac{1}{3} - \dfrac{1}{x + 5}}{x^2 - 4}\cdot \dfrac{3(x + 5)}{3(x + 5)} = \dfrac{\dfrac{3(x + 5)}{3} - \dfrac{3(x + 5)}{x + 5}} {3(x + 5)(x^2 - 4)} = \dfrac{(x + 5) - 3}{3(x + 5)(x^2 - 4)} = \dfrac{x + 2}{3(x + 5)(x^2 - 4)} =$$

$$\dfrac{x + 2}{3(x + 5)(x - 2)(x + 2)} = \dfrac{1}{3(x + 5)(x - 2)}.$$

It's also possible to do this problem by combining $\dfrac{1}{3}$ and $\dfrac{1}{x +
   5}$ over a common denominator.


Example.

$$\dfrac{a}{\left(\dfrac{b}{c}\right)} = \dfrac{a}{\left(\dfrac{b}{c}\right)}\cdot \dfrac{\left(\dfrac{c}{b}\right)}{\left(\dfrac{c}{b}\right)} = \dfrac{\left(\dfrac{ac}{b}\right)}{1} = \dfrac{ac}{b}.$$

The cancellation is only valid if $c
   \ne 0$ : If $c = 0$ , neither the first nor the last expression is defined. This example is useful to remember as a rule:

$$\hbox{ \boxedtext{2}{If you have a fraction with a fraction on the bottom, the bottom fraction ``flips over'' and goes to the top.}}$$

For example,

$$\dfrac{x + 2}{\left(\dfrac{x + 1}{4}\right)} = (x + 2)\cdot \dfrac{4}{x + 1} = \dfrac{4(x + 2)}{x + 1}.$$

However, it's important for you to understand why this rule works.


Example.

$$\dfrac{\left(\dfrac{a}{b}\right)}{c} = \dfrac{\left(\dfrac{a}{b}\right)}{c}\cdot \dfrac{\left(\dfrac{1}{c}\right)}{\left(\dfrac{1}{c}\right)} = \dfrac{\left(\dfrac{a}{bc}\right)}{1} = \dfrac{a}{bc}.$$

The cancellation is only valid if $c
   \ne 0$ : If $c = 0$ , neither the first nor the last expression is defined. This example proves the following rule:

$$\hbox{ \boxedtext{2}{If you have a fraction with a fraction on the top, the top fraction ``drops down''.}}$$

For example,

$$\dfrac{\left(\dfrac{x + 4}{5}\right)}{x - 2} = \dfrac{x + 4}{5(x - 2)}.$$

(I've been wrapping fractions in the top and bottom of big fractions in parentheses to make it clearer what the top and bottom are, but I'll often omit these parentheses to save writing.)


Example. Simplify $\dfrac{x - 3}{\dfrac{1}{9} - \dfrac{1}{x^2}}$ .

$$\dfrac{x - 3}{\dfrac{1}{9} - \dfrac{1}{x^2}} = \dfrac{x - 3}{\dfrac{1}{9} - \dfrac{1}{x^2}}\cdot \dfrac{9x^2}{9x^2} = \dfrac{9x^2(x - 3)}{x^2 - 9} = \dfrac{9x^2(x - 3)}{(x - 3)(x + 3)} = \dfrac{9x^2}{x + 3}.\quad\halmos$$


Example. Simplify $\dfrac{\dfrac{1}{5} - \dfrac{1}{x^2 - 4}} {\dfrac{1}{x - 2} -
   \dfrac{1}{x + 2}}$ .

$$\dfrac{\dfrac{1}{5} - \dfrac{1}{x^2 - 4}} {\dfrac{1}{x - 2} - \dfrac{1}{x + 2}} = \dfrac{\dfrac{1}{5} - \dfrac{1}{x^2 - 4}} {\dfrac{1}{x - 2} - \dfrac{1}{x + 2}}\cdot \dfrac{5(x^2 - 4)}{5(x^2 - 4)} = \dfrac{\dfrac{5(x^2 - 4)}{5} - \dfrac{5(x^2 - 4)}{x^2 - 4}} {\dfrac{x^2 - 4}{x - 2} - \dfrac{x^2 - 4}{x + 2}} =$$

$$\dfrac{(x^2 - 4) - 5}{\dfrac{(x - 2)(x + 2)}{x - 2} - \dfrac{(x - 2)(x + 2)}{x + 2}} = \dfrac{x^2 - 9}{(x + 2) - (x - 2)} = \dfrac{(x - 3)(x + 3)}{4}.\quad\halmos$$


In many cases, the best way to simplify a fraction is to factor the top and bottom and cancel common factors. Remember that the cancellation is only valid if the thing you're cancelling is not 0.


Example. Simplify $\dfrac{a^4 - 1}{a^4 + 2a^2 + 1}$ .

$$\dfrac{a^4 - 1}{a^4 + 2a^2 + 1} = \dfrac{(a^2 - 1)(a^2 + 1)}{(a^2 + 1)^2} = \dfrac{a^2 - 1}{a^2 + 1}.\quad\halmos$$


Example. Simplify $\dfrac{x^3 - 5x^2 + 6x}{x^4 - 4x^3 + 4x^2}$ .

$$\dfrac{x^3 - 5x^2 + 6x}{x^4 - 4x^3 + 4x^2} = \dfrac{x(x^2 - 5x + 6)}{x^2(x^2 - 4x + 4)} = \dfrac{x(x - 2)(x - 3)}{x^2(x - 2)^2} = \dfrac{x - 3}{x(x - 2)}.\quad\halmos$$


Example. Simplify $\dfrac{x^3 - 5x^2}{x^3 - 4x^2 - 5x}$ .

$$\dfrac{x^3 - 5x^2}{x^3 - 4x^2 - 5x} = \dfrac{x^2(x - 5)}{x(x^2 - 4x - 5)} = \dfrac{x^2(x - 5)}{x(x - 5)(x + 1)} = \dfrac{x}{x + 1}.\quad\halmos$$


Example. Simplify $\dfrac{a - b}{a^2 - b^2}$ .

$$\dfrac{a - b}{a^2 - b^2} = \dfrac{a - b}{(a - b)(a + b)} = \dfrac{1}{a + b}.\quad\halmos$$


If you're multiplying two fractions, you can multiply the tops together and multiply the bottoms together:

$$\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{a \cdot c}{b \cdot d}.$$

Try simplifying the result by factoring everything on the top and everything on the bottom, then cancelling common factors.


Example. Simplify $\dfrac{x + 1}{x^2 - 2x + 1}\cdot \dfrac{x^2 - 3x + 2}{x^2 +
   x}$ .

$$\dfrac{x + 1}{x^2 - 2x + 1}\cdot \dfrac{x^2 - 3x + 2}{x^2 + x} = \dfrac{x + 1}{(x - 1)^2}\cdot \dfrac{(x - 1)(x - 2)}{x(x + 1)} = \dfrac{x - 2}{x(x - 1)}.\quad\halmos$$


If you have a big fraction with single fractions on the top and the bottom, you can use the earlier rule to "flip over" the fraction on the bottom and bring it to the top. Then the problem is a multiplication problem like the one I just did, which you can handle by factoring and cancelling.


Example. Simplify $\dfrac{\dfrac{x^2 + x}{x^2 - 2x - 3}} {\dfrac{x^3 + 5x^2}{x^2 +
   2x + 1}}$ .

$$\dfrac{\dfrac{x^2 + x}{x^2 - 2x - 3}} {\dfrac{x^3 + 5x^2}{x^2 + 2x + 1}} = \dfrac{x^2 + x}{x^2 - 2x - 3}\cdot \dfrac{x^2 + 2x + 1}{x^3 + 5x^2} = \dfrac{x(x + 1)}{(x - 3)(x + 1)}\cdot \dfrac{(x + 1)^2}{x^2(x + 5)} = \dfrac{(x + 1)^2}{x(x - 3)(x + 5)}.\quad\halmos$$


Example. Simplify $\dfrac{\dfrac{x^2 - xy}{x^2y + xy^2}} {\dfrac{x^2 - 2xy +
   y^2}{x^2y^2}}$ .

$$\dfrac{\dfrac{x^2 - xy}{x^2y + xy^2}} {\dfrac{x^2 - 2xy + y^2}{x^2y^2}} = \dfrac{x^2 - xy}{x^2y + xy^2}\cdot \dfrac{x^2y^2}{x^2 - 2xy + y^2} = \dfrac{x(x - y)}{xy(x + y)}\cdot \dfrac{x^2y^2}{(x - y)^2} = \dfrac{x^2y}{(x + y)(x - y)}.\quad\halmos$$


You can add fractions by finding a common denominator.

1. Factor the denominators of the given fractions.

2. Multiply each fraction by things of the form

$$\dfrac{(\hbox{junk})}{(\hbox{junk})}.$$

Choose $(\hbox{junk})$ so that all resulting fractions have the same denominator.


Example.

$$\dfrac{2}{10} + \dfrac{7}{15} = \dfrac{2}{2 \cdot 5} + \dfrac{7}{3\cdot 5} = \ldots.$$

The first fraction is missing a "3" that the second fraction has, so I need to multiply the first fraction by $\dfrac{3}{3}$ .

The second fraction is missing a "2" that the first fraction has, so I need to multiply the second fraction by $\dfrac{2}{2}$ .

Since both fractions have a "5" on the bottom, I don't need to do anything with that.

$$\dfrac{2}{10} + \dfrac{7}{15} = \dfrac{2}{2 \cdot 5} + \dfrac{7}{3\cdot 5} = \dfrac{3}{3}\cdot \dfrac{2}{2 \cdot 5} + \dfrac{2}{2}\cdot \dfrac{7}{3\cdot 5} = \dfrac{6}{30} + \dfrac{14}{30} = \dfrac{20}{30} = \dfrac{2}{3}.$$

You may have seen this done by cross-multiplying:

$$\dfrac{2}{10} + \dfrac{7}{15} = \dfrac{30 + 70}{150} = \dfrac{2}{3}.$$

Cross-multiplying is okay for simple fractions, but it will make things hard to simplify if the fractions are complicated. I don't recommend that you use it.


Example. Combine and simplify: $\dfrac{x + 3}{x^2 - 1} + \dfrac{5}{x^2 - x
   - 2}$ .

If I try to do this by cross-multiplying, here's what I'd get:

$$\dfrac{x + 3}{x^2 - 1} + \dfrac{5}{x^2 - x - 2} = \dfrac{(x + 3)(x^2 - x - 2) + 5(x^2 - 1)}{(x^2 - 1)(x^2 - x - 2)} =$$

$$\dfrac{(x^3 - x^2 - 2x + 3x^2 - 3x - 6) + (5x^2 - 5)} {(x^2 - 1)(x^2 - x - 2)} = \dfrac{x^3 + 7x^2 - 5x - 11}{(x^2 - 1)(x^2 - x - 2)}.$$

In the first place, I had a tedious multiplication to do on the top of the fraction. It's very easy to make a mistake.

In the second place, the result isn't simplified; it turns out that

$$x^3 + 7x^2 - 5x - 11 = (x + 1)(x^2 + 6x - 11).$$

Would you have seen that? This cubic doesn't factor by any easy factoring method, such as factoring by grouping. And you have to do this factoring, because the factor $x
   + 1$ will cancel.

Here's how the problem looks if I find a common denominator:

$$\dfrac{x + 3}{x^2 - 1} + \dfrac{5}{x^2 - x - 2} = \dfrac{x + 3}{(x - 1)(x + 1)} + \dfrac{5}{(x - 2)(x + 1)} = \ldots$$

The first fraction is missing an "$x - 2$ " that the second fraction has, so I need to multiply the first fraction by $\dfrac{x - 2}{x - 2}$ .

The second fraction is missing an "$x - 1$ " that the first fraction has, so I need to multiply the second fraction by $\dfrac{x - 1}{x - 1}$ .

Both fractions have an "$x + 1$ " on the bottom, so I don't do anything with that.

$$\dfrac{x + 3}{x^2 - 1} + \dfrac{5}{x^2 - x - 2} = \dfrac{x + 3}{(x - 1)(x + 1)} + \dfrac{5}{(x - 2)(x + 1)} = \dfrac{x - 2}{x - 2}\cdot \dfrac{x + 3}{(x - 1)(x + 1)} + \dfrac{x + 3}{x + 3}\cdot \dfrac{5}{(x - 2)(x + 1)} =$$

$$\dfrac{(x^2 + x - 6) + (5x + 15)}{(x + 3)(x - 2)(x + 1)} = \dfrac{x^2 + 6x - 11}{(x + 3)(x - 2)(x + 1)}.\quad\halmos$$


Example. Combine and simplify: $\dfrac{2}{x^2 - 3x - 4} - \dfrac{5}{x^2 -
   1}$ .

$$\dfrac{2}{x^2 - 3x - 4} - \dfrac{5}{x^2 - 1} = \dfrac{2}{(x - 4)(x + 1)} - \dfrac{5}{(x - 1)(x + 1)} = \dfrac{x - 1}{x - 1}\cdot \dfrac{2}{(x - 4)(x + 1)} - \dfrac{x - 4}{x - 4}\cdot \dfrac{5}{(x - 1)(x + 1)} =$$

$$\dfrac{2(x - 1)}{(x - 1)(x + 1)(x - 4)} - \dfrac{5(x - 4)}{(x - 1)(x + 1)(x - 4)} = \dfrac{2(x - 1) - 5(x - 4)}{(x - 1)(x + 1)(x - 4)} = \dfrac{2x - 2 - 5x + 20}{(x - 1)(x + 1)(x - 4)} =$$

$$\dfrac{-3x + 18}{(x - 1)(x + 1)(x - 4)} = \dfrac{-3(x - 6)}{(x - 1)(x + 1)(x - 4)}.\quad\halmos$$


Example. Combine and simplify: $\dfrac{1}{x^2 + x} + \dfrac{1}{x^2 + 3x +
   2} - \dfrac{2}{x^2 + 2x}$ .

$$\dfrac{1}{x^2 + x} + \dfrac{1}{x^2 + 3x + 2} - \dfrac{2}{x^2 + 2x} = \dfrac{1}{x(x + 1)} + \dfrac{1}{(x + 1)(x + 2)} - \dfrac{2}{x(x + 2)} =$$

$$\dfrac{x + 2}{x + 2}\cdot \dfrac{1}{x(x + 1)} + \dfrac{x}{x}\cdot \dfrac{1}{(x + 1)(x + 2)} - \dfrac{x + 1}{x + 1}\cdot \dfrac{2}{x(x + 2)} = \dfrac{(x + 2) + x - 2(x + 1)}{x(x + 1)(x + 2)} = 0.\quad\halmos$$


Example. Combine and simplify: $\dfrac{18}{x^2 + x - 2} - \dfrac{12}{x^2 -
   1}$ .

$$\dfrac{18}{x^2 + x - 2} - \dfrac{12}{x^2 - 1} = \dfrac{18}{(x + 2)(x - 1)} - \dfrac{12}{(x - 1)(x + 1)} = \dfrac{x + 1}{x + 1}\cdot \dfrac{18}{(x + 2)(x - 1)} - \dfrac{x + 2}{x + 2}\cdot \dfrac{12}{(x - 1)(x + 1)} =$$

$$\dfrac{18(x + 1)}{(x + 1)(x + 2)(x - 1)} - \dfrac{12(x + 2)}{(x + 2)(x - 1)(x + 1)} = \dfrac{18(x + 1) - 12(x + 2)}{(x + 1)(x + 2)(x - 1)} = \dfrac{18x + 18 - 12x - 24}{(x + 1)(x + 2)(x - 1)} =$$

$$\dfrac{6x - 6}{(x + 1)(x + 2)(x - 1)} = \dfrac{6(x - 1)}{(x + 1)(x + 2)(x - 1)} = \dfrac{6}{(x + 1)(x + 2)}.\quad\halmos$$


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