Explainer: Simplifying Rational Functions

In this explainer, we will learn how to simplify rational functions and how to find their domains.

There are two skills in this explainer that are intrinsically linked but can be approached independently. The first is the ability to state the domains of rational functions, and the second is the ability to simplify them. It is important when asked to simplify a rational function that we think about the domain of the function before we simplify it.

Remember, to determine the domain of a rational function, we need to determine any values where the denominator is nonzero. If the function contains multiple rational expressions, we need to determine any points that would result in a denominator of zero. Consider the function 𝑓(π‘₯)=π‘₯+4π‘₯+4π‘₯βˆ’4.

We can factor the numerator and denominator of the function to find that 𝑓(π‘₯)=(π‘₯+2)(π‘₯+2)(π‘₯+2)(π‘₯βˆ’2).

At this point, despite there being a common factor on the top and bottom of the expression, we should not simplify. First, we need to establish if there are any inputs that would result in a denominator of zero. If we solve the equation (π‘₯+2)(π‘₯βˆ’2)=0,

we get π‘₯=2 and π‘₯=βˆ’2. This means that the function is undefined at these points and, therefore, has a domain of β„βˆ’{βˆ’2,2}. For values of π‘₯ not equal to 2 or βˆ’2, we can then simplify the function by canceling (π‘₯+2) from the top and bottom of the expression, leaving us with 𝑓(π‘₯)=π‘₯+2π‘₯βˆ’2,

with domain β„βˆ’{βˆ’2,2}. As we said at the start, it is important to determine the domain of the function before we simplify. Had we found the domain subsequently, we would have written that it was β„βˆ’{2}, having lost in the cancellation process that the function is also undefined when π‘₯=βˆ’2.

Let us look at an example.

Example 1: Simplifying Rational Functions and Stating Their Domains

Simplify the function 𝑓(π‘₯)=7π‘₯+43π‘₯+67π‘₯+50π‘₯+7,

and find its domain.

Answer

We need to start by finding the function’s domain. To do this, we need to determine the zeros of the denominator of the rational function. We need to solve the equation 7π‘₯+50π‘₯+7=0.

We can factor the expression on the left-hand side to get (7π‘₯+1)(π‘₯+7)=0,

which has solutions of π‘₯=βˆ’17 and π‘₯=βˆ’7. Therefore, the domain of the rational function is β„βˆ’ο¬βˆ’17,βˆ’7.

To simplify the expression, we need to factor the numerator and denominator to get 𝑓(π‘₯)=(7π‘₯+1)(π‘₯+6)(7π‘₯+1)(π‘₯+7).

We can now cancel 7π‘₯+1 from the top and the bottom of the expression: 𝑓(π‘₯)=(7π‘₯+1)(π‘₯+6)(7π‘₯+1)(π‘₯+7).

We can now see that 𝑓(π‘₯)=π‘₯+6π‘₯+7,

with domain β„βˆ’ο¬βˆ’17,βˆ’7.

With some questions, we may need to apply one of the four operations: addition, subtraction, multiplication, or division. With questions like this, we take a very similar approach to the previous example but have to work a little harder to simplify the function into a single rational expression. To recap, to add or subtract two rational expressions, they each must have the same denominator; to multiply two rational expressions, you multiply the numerators and multiply the denominators; and to divide a rational expression by another rational expression, you flip the second one and then multiply (i.e., multiply by the reciprocal). Let us look at an example of each of these operations now.

Example 2: Stating the Domain of a Function That Is a Sum of Two Rational Expressions and Simplifying It

Simplify the function 𝑛(π‘₯)=π‘₯π‘₯+9π‘₯+π‘₯+9π‘₯βˆ’81,

and determine its domain.

Answer

Before we do any simplification of the two rational expressions, we need to determine the domain of the function. To do this, we need to determine where each of the denominators will be equal to zero. For the first expression, we need to solve the equation π‘₯+9π‘₯=0.

Factoring, we get π‘₯(π‘₯+9)=0,

which has solutions π‘₯=0 and π‘₯=βˆ’9. For the second expression we get the equation π‘₯βˆ’81=0,

which factors to (π‘₯+9)(π‘₯βˆ’9)=0

and has solutions π‘₯=9 and π‘₯=βˆ’9. Therefore, the function is undefined when π‘₯=0,9,βˆ’9, so the function has domain β„βˆ’{0,9,βˆ’9}. At this point, we can now simplify the function. If we rewrite each of the expressions with their factored denominators, we have 𝑛(π‘₯)=π‘₯π‘₯(π‘₯+9)+π‘₯+9(π‘₯βˆ’9)(π‘₯+9).

To combine the two expressions, we need to ensure that they have a common denominator. If we multiply the top and bottom of the first expression by π‘₯βˆ’9, we get 𝑛(π‘₯)=π‘₯(π‘₯βˆ’9)π‘₯(π‘₯βˆ’9)(π‘₯+9)+π‘₯+9(π‘₯βˆ’9)(π‘₯+9).

We can now cancel the π‘₯ from the top and bottom of the first expression and then we can combine the two expressions to get 𝑛(π‘₯)=π‘₯βˆ’9+π‘₯+9(π‘₯βˆ’9)(π‘₯+9).

Simplifying, we find that 𝑛(π‘₯)=2π‘₯(π‘₯βˆ’9)(π‘₯+9),

with domain β„βˆ’{0,9,βˆ’9}.

Example 3: Stating the Domain of a Function That Is a Difference of Two Rational Expressions and Simplifying It

Simplify the function 𝑛(π‘₯)=5π‘₯π‘₯βˆ’4βˆ’π‘₯+4π‘₯βˆ’16,

and determine its domain.

Answer

Before we do any simplification of the two rational expressions, we need to determine the domain of the function. To do this, we need to determine where each of the denominators will be equal to zero. For the first expression, we need to solve the equation π‘₯βˆ’4=0,

which has a solution of π‘₯=4. For the second expression, we get the equation π‘₯βˆ’16=0,

which factors to (π‘₯+4)(π‘₯βˆ’4)=0

and has solutions π‘₯=4 and π‘₯=βˆ’4. Therefore, the function is undefined when π‘₯=4 and π‘₯=βˆ’4 and thus has domain β„βˆ’{4,βˆ’4}. At this point, we can simplify the function. If we rewrite each of the expressions with their factored denominators, we have 𝑛(π‘₯)=5π‘₯π‘₯βˆ’4βˆ’π‘₯+4(π‘₯+4)(π‘₯βˆ’4).

To combine the two expressions, we need to ensure that they have a common denominator. If we multiply the top and bottom of the first expression by π‘₯+4, we get 𝑛(π‘₯)=5π‘₯(π‘₯+4)(π‘₯βˆ’4)(π‘₯+4)βˆ’π‘₯+4(π‘₯βˆ’4)(π‘₯+4).

We can now combine the two expressions to get 𝑛(π‘₯)=5π‘₯(π‘₯+4)βˆ’(π‘₯+4)(π‘₯βˆ’4)(π‘₯+4).

Since both of the terms in the numerator contain a common factor of π‘₯+4, we can factor this out as follows: 𝑛(π‘₯)=(π‘₯+4)(5π‘₯βˆ’1)(π‘₯βˆ’4)(π‘₯+4).

We can now cancel the common factor in the numerator and denominator to get 𝑛(π‘₯)=5π‘₯βˆ’1π‘₯βˆ’4,

with domain β„βˆ’{4,βˆ’4}.

Having looked at examples of a sum and then a difference of two rational expressions, let us now look at examples of a product and then a quotient of two rational expressions.

Example 4: Stating the Domain of a Function That Is a Product of Two Rational Expressions and Simplifying It

Simplify the function 𝑛(π‘₯)=π‘₯+5π‘₯+9π‘₯+20Γ—π‘₯+15π‘₯+547π‘₯+69π‘₯+54,

and determine its domain.

Answer

Before we do any simplification of the two rational expressions, we need to determine the domain of the function. To do this, we need to determine where each of the denominators will be equal to zero. For the first expression, we need to solve the equation π‘₯+9π‘₯+20=0.

We can factor the expression to get the equation (π‘₯+5)(π‘₯+4)=0,

which has solutions π‘₯=βˆ’5 and π‘₯=βˆ’4. For the second expression, we get the equation 7π‘₯+69π‘₯+54=0,

which factors to (7π‘₯+6)(π‘₯+9)=0

and has solutions π‘₯=βˆ’67 and π‘₯=βˆ’9. Therefore, the function is undefined at each of these points and thus has domain β„βˆ’ο¬βˆ’9,βˆ’5,βˆ’4,βˆ’67. At this point, we can simplify the function. If we multiply the numerators and denominators of the two expressions and write each of the quadratics in factored form, we get 𝑛(π‘₯)=(π‘₯+5)(π‘₯+9)(π‘₯+6)(π‘₯+5)(π‘₯+4)(7π‘₯+6)(π‘₯+9).

If we now cancel any common factors from the top and bottom of the expression, we get 𝑛(π‘₯)=(π‘₯+5)(π‘₯+9)(π‘₯+6)(π‘₯+5)(π‘₯+4)(7π‘₯+6)(π‘₯+9).

This leaves us with the function 𝑛(π‘₯)=π‘₯+6(π‘₯+4)(7π‘₯+6),

with domain β„βˆ’ο¬βˆ’9,βˆ’5,βˆ’4,βˆ’67.

Example 5: Stating the Domain of a Function That Is a Quotient of Two Rational Expressions

Determine the domain of the function 𝑛(π‘₯)=π‘₯βˆ’648π‘₯+7π‘₯Γ·9π‘₯βˆ’117π‘₯+36064π‘₯βˆ’49.

Answer

With the quotient of two rational expressions, to determine the domain, there are a couple of things that we need to determine. When the denominators of the two rational expressions are zero, and when the divisor is zero, in this case where the numerator of the rational expression is zero, we need to solve three equations to find the domain of the function: 8π‘₯+7π‘₯=0,64π‘₯βˆ’49=0,9π‘₯βˆ’117π‘₯+360=0.

The first equation can be factored to give us π‘₯(8π‘₯+7)=0,

which has solutions of π‘₯=0 and π‘₯=βˆ’78. The second equation can be factored to give us (8π‘₯βˆ’7)(8π‘₯+7)=0,

which has solutions of π‘₯=78 and π‘₯=βˆ’78. Finally, the third equation can be factored to give us 9((π‘₯βˆ’5)(π‘₯βˆ’8))=0,

which has solutions of π‘₯=5 and π‘₯=8. Therefore, the domain of the function is β„βˆ’ο¬βˆ’78,0,78,5,8.

As an additional piece of information, had we been asked to simplify the function, having now found the domain, we could have done this by multiplying by the reciprocal of the divisor and writing each of the quadratics in factored form. This would give us 𝑛(π‘₯)=(π‘₯βˆ’8)(π‘₯+8)π‘₯(8π‘₯+7)Γ—(8π‘₯+7)(8π‘₯βˆ’7)9(π‘₯βˆ’5)(π‘₯βˆ’8).

At this point, we can cross-cancel, giving us 𝑛(π‘₯)=(π‘₯βˆ’8)(π‘₯+8)π‘₯(8π‘₯+7)Γ—(8π‘₯+7)(8π‘₯βˆ’7)9(π‘₯βˆ’5)(π‘₯βˆ’8).

We are then left with the function 𝑛(π‘₯)=(π‘₯+8)(8π‘₯βˆ’7)9π‘₯(π‘₯βˆ’5)

with domain β„βˆ’ο¬βˆ’78,0,78,5,8.

Key Points

  1. When simplifying rational functions, always determine the domain first; otherwise, in the simplification process, you may lose information about the function.
  2. When the function is a sum, difference, product, or quotient of two rational expressions, determine the domain first by identifying any undefined points and then combine the expressions and simplify using standard methods.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.