# Rational Expressions and Their Domain

The definition and domain of rational expressions are examined. Exercises with their solutions are also included.

## Definition of Rational Expressions

A rational expression is the quotient of two polynomials.

### Example 1 : Rational Expressions

1. $\dfrac{x-1}{x+2}$

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

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

4. $\dfrac{x^2+4}{x^3+2x^2-3x}$

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

## Domain of Rational Expressions

Being a ratio, rational expressions are undefined if a division by $0$ occurs. Hence, the domain of a rational expression excludes all values that make the denominator equal to $0$.
To find the domain of a rational function of the form $\dfrac{N(x)}{D(x)}$, is to
1 - solve the equation $D(x) = 0$
2 - write the domain as the set of all the real numbers excluding the solutions to the equation $D(x) = 0$.
Note that the domain may be written in three different ways:
a) all set of real numbers except the solutions of the equation $D(x) = 0$
b) using intervals
c) using inequalities

### Example 2

What is the domain of each of the rational expression in example 1 above?

### Solutions

1. The domain of the rational expression $\dfrac{x-1}{x+2}$ is found by
1 - solving the equation $x + 2 = 0$ , whose solution is $x = -2$
2 - exclude the solution $x = - 2$ from the set of all real numbers.
Hence the domain of the given rational expression may be given by any of the following
a) the set of all real numbers except $x = - 2$.
b) using interval notation, the domain is written as: $\quad (-\infty , -2) \cup (-2 , +\infty)$
c) using inequalities, the domain is written as: $\quad x \lt - 2 \; \quad \text{ or } \quad \; x \gt -2$

2. The domain of the rational expression $\dfrac{2 x+1}{(x-2)(x+3)}$ is found by
1 - solving the equation $(x-2)(x+3) = 0$ , whose solutions are $x = 2$ and $x = 3$
2 - exclude the solutions $x = 2$ and $x = 3$ from the set of all real numbers.
The domain is given by any of the following
a) the set of all real numbers except $x = 2$ and $x = 3$.
b) using interval notation, the domain is written as: $\quad (-\infty , 2) \cup (2 , 3) \cup (3 , +\infty)$
c) Using inequalities, the domain is written as: $\quad x \lt 2 \; \quad \text{ or } \quad \; 2 \lt x \lt 3 \; \quad \text{ or } \quad \; x \gt 3$

3. The domain of the rational expression $\dfrac{-x+3}{x^2+1}$ is found
1 - solving the equation $x^2+1 = 0$ which does not have any real solutions.
The domain is given by any of the following
a) the set of all real numbers.
b) using interval notation, the domain is written as: $\quad (-\infty , +\infty)$

4. The domain of the rational expression $\dfrac{x^2+4}{x^3+2x^2-3x}$ is found
1 - solving the equation $x^3+2x^2-3x= 0$
First factor the left hand side of the equation
$x^3+2x^2-3x \\ = x(x^2 + 2 x - 3) \\ = x(x - 1)(x + 3)$
then solve
$x(x - 1)(x + 3) = 0$ whose solutions are: $x = -3$ , $x = 0$ and $x = 1$
The domain is given by any of the following
a) the set of all real numbers except $x = -3$ , $x = 0$ and $x = 1$
b) using interval notation, the domain is written as: $\quad (-\infty , -3) \cup (-3 , 0) \cup (0 , 1) \cup (1 , \infty)$
c) inequatlity notation: $\quad x \lt - 3 \quad \text{ or } \quad -3\lt x \lt 0 \quad \text{ or } \quad 0\lt x \lt 1 \quad \text{ or } \quad x \gt 1$

5. The domain of the rational expression $\dfrac{x^2+2x+1}{x^2+x-1}$ is found by solving the equation $x^2+x-1 = 0$
$\Delta = b^2 - 4 a c = (1)^2 - 4 (1)(-1) = 5$
Solutions to the above equation are
$x_1 = \dfrac{-b-\sqrt \Delta }{2 a} = \dfrac{-1-\sqrt 5}{2}$
and
$x_2 = \dfrac{-b+\sqrt \Delta }{2 a} = \dfrac{-1+\sqrt 5}{2}$
The domain is given by any of the following
a) the set of all real numbers except $x = \dfrac{-1-\sqrt 5}{2}$ and $x = \dfrac{-1+\sqrt 5}{2}$

b) using interval notation, the domain is written as: $\quad (-\infty , \dfrac{-1-\sqrt 5}{2}) \cup (\dfrac{-1-\sqrt 5}{2}, \dfrac{-1+\sqrt 5}{2}) \cup (\dfrac{-1+\sqrt 5}{2} , \infty)$

c) inequatlity notation: $\quad x \lt \dfrac{-1-\sqrt 5}{2} \quad \text{ or } \quad \dfrac{-1-\sqrt 5}{2} \lt x \lt \dfrac{-1+\sqrt 5}{2} \quad \text{ or } \quad x \gt \dfrac{-1+\sqrt 5}{2}$

## Exercises

Find the domain of each of the Rational Expressions given below.

1. $\dfrac{x+9}{x-10}$

2. $\dfrac{1}{(x-9)(x+1)}$

3. $\dfrac{1}{x^3+1}$

4. $\dfrac{-x+7}{x^4 - 16}$

## Solutions

Solution to Exercise 1
The domain of $\dfrac{x+9}{x-10}$ is found by first solving the equation $x-10 = 0$ whose solution is $x = 10$.
The domain is given by any of the following
a) the set of all real numbers except $x = 10$
b) using interval notation, the domain is written as: $\quad (-\infty , 10) \cup (10, \infty)$
c) inequatlity notation: $\quad x \lt 10 \quad \text{ or } \quad x \gt 10$

Solution to Exercise 2
The domain of $\dfrac{1}{(x-9)(x+1)}$ is found by first solving the equation $(x-9)(x+1)$ whose solutions are $x = -1$ and $x = 9$.
The domain is given by any of the following
a) the set of all real numbers except $x = -1$ and $x = 9$
b) using interval notation, the domain is written as: $\quad (-\infty , -1) \cup (-1, 9 ) \cup (9, \infty)$
c) inequatlity notation: $\quad x \lt -1 \quad \text{ or } \quad -1 \lt x \lt 9 \quad \text{ or } \quad x \gt 9$

Solution to Exercise 3
The domain of $\dfrac{1}{x^3+1}$ is found by first solving the equation $x^3 + 1 = 0$ whose solution is $x = -1$.
The domain is given by any of the following
a) the set of all real numbers except $x = -1$.
b) using interval notation, the domain is written as: $\quad (-\infty , -1) \cup (-1, \infty)$
c) inequatlity notation: $\quad x \lt -1 \quad \text{ or } \quad x \gt -1$

Solution to Exercise 4
The domain of $\dfrac{-x+7}{x^4 - 16}$ is found by first solving the equation $x^4 - 16 = 0$ which may be solved by factoring as follows:
$x^4 - 16 = ( x^2 - 4 ) ( x^2 + 4 ) = (x - 2)(x + 2) ( x^2 + 4 )$
The real solutions to the equation $x^4 - 16 = 0$ are $x = - 2$ and $x = 2$
The domain is given by any of the following
a) the set of all real numbers except $x = - 2$ and $x = 2$.
b) using interval notation, the domain is written as: $\quad (-\infty , -2) \cup (-2, 2 ) \cup (2, \infty)$
c) inequatlity notation: $\quad x \lt -2 \quad \text{ or } \quad -2 \lt x \lt 2 \quad \text{ or } \quad x \gt 2$