Fractions
An algebraic fraction (also called a rational expression) is a fraction whose numerator and denominator are polynomials, such as $\dfrac{x^2-1}{x+3}$.
We manipulate algebraic fractions the same way that we manipulate fractions in arithmetic.
Quick Reference
| Operation | Formula | Notes |
|---|---|---|
| Equivalent fractions | $\dfrac{A}{B} = \dfrac{AC}{BC}$ | $C \neq 0$ |
| Simplification | Cancel common factors from numerator and denominator | Cancelled factors must be nonzero |
| Multiplication | $\dfrac{A}{B} \cdot \dfrac{C}{D} = \dfrac{AC}{BD}$ | $B \neq 0$, $D \neq 0$ |
| Division | $\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{AD}{BC}$ | $B \neq 0$, $C \neq 0$, $D \neq 0$ |
| Addition/Subtraction (same denominator) | $\dfrac{A}{B} \pm \dfrac{C}{B} = \dfrac{A \pm C}{B}$ | $B \neq 0$ |
| Addition/Subtraction (different denominators) | $\dfrac{A}{B} \pm \dfrac{C}{D} = \dfrac{AD \pm BC}{BD}$ | $B \neq 0$, $D \neq 0$ |
| Compound fractions (Method 2) | Multiply numerator and denominator by the LCD | Clears all inner fractions at once |
Simplifying Fractions
One fundamental rule for manipulation of fractions is: If we multiply or divide both the numerator or denominator by the same quantity, the value of the fraction will not change provided that this quantity is nonzero, namely
$ \frac{A}{B} = \frac{AC}{BC}, \qquad \frac{A}{B} = \frac{\dfrac{A}{C}}{\dfrac{B}{C}} \qquad (C \neq 0) $
For example, multiplying both the numerator and denominator of $\dfrac{x-1}{x+2}$ by $(x-3)$ gives the equivalent fraction
$ \frac{x-1}{x+2} = \frac{(x-1)(x-3)}{(x+2)(x-3)} $
provided $x-3 \neq 0$, that is, provided $x \neq 3$.
Conversely, if we factor the numerator and denominator of a fraction, we can cancel common factors to simplify the fraction (again, provided those common factors are nonzero). For example:
$ \frac{4}{12} = \frac{\cancel{4}}{\cancel{4} \times 3} = \frac{1}{3} $$ \frac{x^{2}-5x+6}{x^{2}-4x+3} = \frac{(x-2)\cancel{(x-3)}}{\cancel{(x-3)}(x-1)} = \frac{x-2}{x-1} \qquad (x \neq 3) $
Example 1. Simplify $\dfrac{x^{2}+x-12}{3-x}$.
Solution
Factor the numerator, then observe that the denominator $3-x = -(x-3)$: $ \begin{aligned} \frac{x^{2}+x-12}{3-x} &= \frac{(x+4)(x-3)}{3-x} \\[6pt] &= \frac{(x+4)\cancel{(x-3)}}{-\cancel{(x-3)}} \\[6pt] &= -(x+4) \\[6pt] &= -x-4 \end{aligned} $ The key observation is that $3-x = -(x-3)$, which allows the cancellation. The result holds for $x \neq 3$.
Multiplying and Dividing Fractions
To multiply or divide fractions, use the following rules:
Multiplication Rule:
$\frac{A}{B} \cdot \frac{C}{D} = \frac{A \cdot C}{B \cdot D} \qquad (B \neq 0,\ D \neq 0)$To multiply two fractions, multiply their numerators together and multiply their denominators together.
Division Rule:
$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC} \qquad (B \neq 0,\ C \neq 0,\ D \neq 0)$To divide by a fraction, invert the divisor and multiply. In other words, "flip" the second fraction and then multiply.
Example 2. Perform the following operations and simplify the results.
- $\dfrac{x^{2}-2x-3}{x^{2}+6x+9}\cdot\dfrac{4x+12}{x+1}$
- $\dfrac{x^{2}-6x+8}{x^{2}-x-6}\div\dfrac{x^{2}-x-12}{4x-12}$
Solution
(a) Factor each polynomial completely, then cancel common factors before multiplying: $ \begin{aligned} \frac{x^{2}-2x-3}{x^{2}+6x+9}\cdot\frac{4x+12}{x+1} &= \frac{(x-3)\bcancel{(x+1)}}{(x+3)^{\cancel{2}}} \cdot \frac{4\cancel{(x+3)}}{\bcancel{x+1}} \\[6pt] &= \frac{4(x-3)}{x+3} \end{aligned} $ provided $x \neq -1$.(b) Rewrite the division as multiplication by the reciprocal, then factor and cancel: $ \begin{aligned} \frac{x^{2}-6x+8}{x^{2}-x-6} \div \frac{x^{2}-x-12}{4x-12} &= \frac{x^{2}-6x+8}{x^{2}-x-6} \cdot \frac{4x-12}{x^{2}-x-12} \\[6pt] &= \frac{\bcancel{(x-4)}(x-2)}{\cancel{(x-3)}(x+2)} \cdot \frac{4\cancel{(x-3)}}{\bcancel{(x-4)}(x+3)} \\[6pt] &= \frac{4(x-2)}{(x+2)(x+3)} \end{aligned} $ provided $x \neq 4$ and $x \neq 3$.
Adding and Subtracting Fractions
If the denominators of the fractions are the same, simply add or subtract the numerators:
$ \frac{A}{B} \pm \frac{C}{B} = \frac{A \pm C}{B} $
If the denominators are different, we must first find a common denominator. One straightforward method is to multiply the numerator and denominator of each fraction by the denominator of the other:
$ \begin{aligned} \frac{A}{B} \pm \frac{C}{D} &= \frac{A}{B} \cdot \frac{D}{D} \pm \frac{C}{D} \cdot \frac{B}{B} \\[6pt] &= \frac{AD \pm CB}{BD} \end{aligned} $
For example:
$ \frac{5}{6} + \frac{4}{9} = \frac{5 \cdot 9 + 4 \cdot 6}{6 \cdot 9} = \frac{45 + 24}{54} = \frac{69}{54} = \frac{23}{18} $
To make simplification easier, we often find the least common multiple (LCM) of the denominators, called the least common denominator (LCD), instead of multiplying the denominators together. In the example above, the LCM of 6 and 9 is 18 (since $6 \times 3 = 9 \times 2 = 18$), so we only need to multiply each fraction by a smaller factor:
$ \frac{5}{6} + \frac{4}{9} = \frac{5 \cdot 3}{6 \cdot 3} + \frac{4 \cdot 2}{9 \cdot 2} = \frac{15 + 8}{18} = \frac{23}{18} $
Least Common Multiple (LCM) and Least Common Denominator (LCD)
The least common multiple (LCM) of two or more polynomials is the polynomial of lowest degree with smallest numerical coefficients that is divisible by each of them (that is, dividing it by any one of the polynomials leaves remainder zero).
The LCM of the denominators of a set of fractions is called the least common denominator (LCD).
How to find the LCM of two or more expressions:
- Factor each expression completely.
- Select all distinct factors and assign each the highest exponent with which it appears in any of the expressions.
- Multiply all the selected factors together.
Example 3. Perform the operation and write in simplified form:
$ \frac{4x}{x^{2}-4} + \frac{3x}{x^{2}-5x+6} $Solution
First, factor both denominators to identify the LCD: $ \frac{4x}{x^{2}-4} + \frac{3x}{x^{2}-5x+6} = \frac{4x}{(x-2)(x+2)} + \frac{3x}{(x-3)(x-2)} $ The distinct factors are $(x-2)$, $(x+2)$, and $(x-3)$, each appearing to the first power, so the LCD is $(x-2)(x+2)(x-3)$. Multiply each fraction by the factor it needs to reach the LCD, then add the numerators: $ \begin{aligned} \frac{4x}{(x-2)(x+2)} &+ \frac{3x}{(x-3)(x-2)}\\ &= \frac{4x(x-3)}{(x-2)(x+2)(x-3)} + \frac{3x(x+2)}{(x-3)(x-2)(x+2)} \\[6pt] &= \frac{4x^{2}-12x + 3x^{2}+6x}{(x-3)(x-2)(x+2)} \\[6pt] &= \frac{7x^{2}-6x}{(x-3)(x-2)(x+2)} \\[6pt] &= \frac{x(7x-6)}{(x-3)(x-2)(x+2)} \end{aligned} $
Compound Fractions
A compound fraction is a fraction that has one or more fractions in its numerator, denominator, or both. There are two standard methods for simplifying compound fractions.
Method 1 (Combine then divide): Simplify the numerator and denominator separately, each into a single fraction, then divide the two resulting fractions.
Method 2 (Multiply by LCD): Find the LCD of all the inner fractions appearing anywhere in the expression. Multiply both the numerator and denominator of the compound fraction by that LCD. This clears all inner fractions in one step.
Example 4. Simplify
$ \frac{\dfrac{1}{x^{2}}-4}{2-\dfrac{1}{x}} $Solution
Method 1 (Combine then divide). Combine the numerator over $x^2$ and the denominator over $x$, then divide: $ \begin{aligned} \frac{\dfrac{1}{x^{2}}-4}{2-\dfrac{1}{x}} &= \frac{\dfrac{1-4x^{2}}{x^{2}}}{\dfrac{2x-1}{x}} \\[10pt] &= \frac{1-4x^{2}}{x^{\cancel{2}}} \cdot \frac{\cancel{x}}{2x-1} \\[10pt] &= \frac{1-4x^{2}}{x(2x-1)} \end{aligned} $ Using the Difference of Squares formula $A^{2}-B^{2}=(A-B)(A+B)$ with $A=1$ and $B=2x$, we have $1-4x^{2}=(1-2x)(1+2x)$. Also note that $2x-1=-(1-2x)$, so: $ \begin{aligned} &= \frac{(1-2x)(1+2x)}{x(2x-1)} \\[10pt] &= \frac{\bcancel{(1-2x)}(1+2x)}{-x\bcancel{(1-2x)}} \\[10pt] &= -\frac{1+2x}{x} \end{aligned} $ Method 2 (Multiply by LCD). The LCD of $x^{2}$ and $x$ is $x^{2}$. Multiply the entire compound fraction by $\dfrac{x^2}{x^2}$: $ \begin{aligned} \frac{\dfrac{1}{x^{2}}-4}{2-\dfrac{1}{x}} \cdot \frac{x^{2}}{x^{2}} &= \frac{1-4x^{2}}{2x^{2}-x} \\[10pt] &= \frac{(1-2x)(1+2x)}{x(2x-1)} \\[10pt] &= -\frac{1+2x}{x} \end{aligned} $ Method 3 (Factor as difference of squares directly). This is a clever shortcut specific to this problem's structure. Notice that $ \begin{aligned} \frac{1}{x^{2}}-4 &= \left(\frac{1}{x}\right)^{2}-2^{2} \\[10pt] &= \left(\frac{1}{x}-2\right)\left(\frac{1}{x}+2\right) \end{aligned} $ and that $2-\dfrac{1}{x} = -\left(\dfrac{1}{x}-2\right)$. Therefore: $ \begin{aligned} \frac{\dfrac{1}{x^{2}}-4}{2-\dfrac{1}{x}} &= \frac{\left(\dfrac{1}{x}-2\right)\left(\dfrac{1}{x}+2\right)}{-\left(\dfrac{1}{x}-2\right)} \\[10pt] &= -\left(\frac{1}{x}+2\right) \\[10pt] &= -\frac{1+2x}{x} \end{aligned} $Example 5. Simplify
$ \frac{(1+x)^{1/2}-x(1+x)^{-1/2}}{x+1} $