Rational expressions

A fraction whose numerator and denominator are polynomials is called a rational expression. For example,

${\dfrac{x+1}{x^2-9}}$   and   ${\dfrac{3}{x}}$

are rational expressions. They follow the same rules as fractions of numbers. This page shows how to simplify them, multiply and divide them, and add and subtract them. Factoring is the key step every time.

The domain of a rational expression

We can never divide by ${0}$. So a rational expression is undefined for any value of ${x}$ that makes its denominator ${0}$. The domain is the set of all real numbers we may put in for ${x}$: every real number except those.

Example 1: Find the domain of ${\dfrac{x+1}{x^2-9}}$.

Solution:

Find where the denominator is ${0}$. Factor it:

$\begin{align*}x^2-9&=0\\(x+3)(x-3)&=0\end{align*}$

So the denominator is ${0}$ when ${x=-3}$ or ${x=3}$. The domain is all real numbers except ${-3}$ and ${3}$.

The numerator does not matter here. When ${x=-1}$, the expression is ${\dfrac{0}{-8}=0}$, which is fine.

Simplifying

A rational expression is in lowest terms when its numerator and denominator have no common factor. To simplify:

  1. Factor the numerator and the denominator completely.
  2. Cancel the common factors.

This works because ${\dfrac{AC}{BC}=\dfrac{A}{B}}$ when ${C\ne 0}$.

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

Solution:

Factor the numerator (a difference of squares) and the denominator (a trinomial):

$\begin{align*}&\dfrac{x^2-4}{x^2+x-6}\\&=\dfrac{(x+2)(x-2)}{(x+3)(x-2)}\end{align*}$

Cancel the common factor ${x-2}$:

$\begin{align*}&=\dfrac{x+2}{x+3}\end{align*}$

The two forms are equal for every ${x}$ in the domain of the original expression, that is, for ${x\ne 2}$ and ${x\ne -3}$.

Cancel factors, never terms. A factor is multiplied; a term is added. For example,

${\dfrac{x+3}{3}\ne x}$.

Here ${3}$ is a term of the numerator, not a factor, so it cannot be canceled. Check with ${x=1}$: the left side is ${\dfrac{4}{3}}$, not ${1}$.

Multiplying and dividing

Multiply rational expressions the same way as fractions: multiply the numerators, and multiply the denominators. Factor first, so you can cancel before multiplying out.

To divide, multiply by the reciprocal of the second expression (flip it upside down):

$\dfrac{A}{B}\div\dfrac{C}{D}=\dfrac{A}{B}\cdot\dfrac{D}{C}$

Example 3: Simplify.

(a) $\dfrac{x^2-1}{x^2+2x}\cdot\dfrac{x+2}{x+1}$

(b) ${\dfrac{x^2-9}{4x}\div\dfrac{x+3}{2x^2}}$

Solution:

(a) Factor everything:

$\begin{align*}&\dfrac{x^2-1}{x^2+2x}\cdot\dfrac{x+2}{x+1}\\&=\dfrac{(x+1)(x-1)}{x(x+2)}\cdot\dfrac{x+2}{x+1}\end{align*}$

Cancel the common factors ${x+1}$ and ${x+2}$:

$\begin{align*}&=\dfrac{x-1}{x}\end{align*}$

(b) Multiply by the reciprocal of the second fraction. Then factor:

$\begin{align*}&\dfrac{x^2-9}{4x}\div\dfrac{x+3}{2x^2}\\&=\dfrac{x^2-9}{4x}\cdot\dfrac{2x^2}{x+3}\\&=\dfrac{(x+3)(x-3)}{4x}\cdot\dfrac{2x^2}{x+3}\end{align*}$

Cancel ${x+3}$. Also, ${\dfrac{2x^2}{4x}=\dfrac{x}{2}}$:

$\begin{align*}&=\dfrac{x(x-3)}{2}\end{align*}$

Adding and subtracting

To add or subtract fractions, they need a common denominator. With rational expressions, we use the least common denominator (LCD):

  1. Factor each denominator.
  2. The LCD is the product of every different factor, each raised to the highest power it has in any denominator.
  3. Rewrite each fraction with the LCD. Multiply its numerator and denominator by the missing factors.
  4. Add or subtract the numerators. Keep the LCD.
  5. Simplify, if possible.

Example 4: Simplify.

(a) ${\dfrac{3}{x-2}+\dfrac{5}{x+2}}$

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

Solution:

(a) The LCD is ${(x-2)(x+2)}$. Give each fraction this denominator:

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

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

Now add the numerators, and combine like terms:

$\begin{align*}&\dfrac{3}{x-2}+\dfrac{5}{x+2}\\&=\dfrac{3(x+2)+5(x-2)}{(x-2)(x+2)}\\&=\dfrac{3x+6+5x-10}{(x-2)(x+2)}\\&=\dfrac{8x-4}{(x-2)(x+2)}\end{align*}$

(b) Factor the first denominator: ${x^2-1=(x+1)(x-1)}$. So the LCD is ${(x+1)(x-1)}$. Only the second fraction needs to change:

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

Now subtract the whole numerator ${x-1}$. Use parentheses, so both of its terms change sign:

$\begin{align*}&\dfrac{x}{x^2-1}-\dfrac{1}{x+1}\\&=\dfrac{x-(x-1)}{(x+1)(x-1)}\\&=\dfrac{1}{(x+1)(x-1)}\end{align*}$

Complex fractions

A complex fraction is a fraction with fractions in its numerator or its denominator. To simplify one, multiply its numerator and denominator by the LCD of all the small fractions inside it.

Example 5: Simplify ${\dfrac{\dfrac{1}{x}-\dfrac{1}{2}}{x-2}}$.

Solution:

The small fractions have denominators ${x}$ and ${2}$. Their LCD is ${2x}$. Multiply the numerator and the denominator by ${2x}$:

$\begin{align*}&\dfrac{\dfrac{1}{x}-\dfrac{1}{2}}{x-2}\cdot\dfrac{2x}{2x}\\&=\dfrac{2-x}{2x(x-2)}\end{align*}$

The numerator ${2-x}$ is the negative of ${x-2}$: ${2-x=-(x-2)}$. Cancel:

$\begin{align*}&=\dfrac{-(x-2)}{2x(x-2)}\\&=-\dfrac{1}{2x}\end{align*}$

Rationalizing a numerator

On the page on exponents and radicals, we rationalized denominators. Sometimes we do the same with a numerator instead. This step is used often in calculus. We multiply the numerator and the denominator by the conjugate of the numerator.

Example 6: Rationalize the numerator of ${\dfrac{\sqrt{x+4}-2}{x}}$.

Solution:

The conjugate of ${\sqrt{x+4}-2}$ is ${\sqrt{x+4}+2}$. In the numerator, use ${(a-b)(a+b)=a^2-b^2}$:

$\begin{align*}&\dfrac{\sqrt{x+4}-2}{x}\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\\&=\dfrac{(x+4)-4}{x(\sqrt{x+4}+2)}\\&=\dfrac{x}{x(\sqrt{x+4}+2)}\end{align*}$

Cancel the common factor ${x}$:

$\begin{align*}&=\dfrac{1}{\sqrt{x+4}+2}\end{align*}$

Summary

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