Lesson Explainer: Multiplying and Dividing Rational Functions Mathematics

In this explainer, we will learn how to multiply and divide rational functions.

Let us recall the definition of a rational function.

Definition: Rational Functions

A function ๐‘“โˆถ๐‘‹โŸถ๐‘Œ is called a rational function if it can be written in the form ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ), where ๐‘ and ๐‘ž are polynomial functions and ๐‘ž(๐‘ฅ)โ‰ 0 for all ๐‘ฅโˆˆ๐‘‹.

Recall that the domain ๐‘‹ of a rational function ๐‘“(๐‘ฅ) depends upon the denominator. If ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ), then ๐‘ฅ cannot take values such that ๐‘ž(๐‘ฅ)=0 because then we would be dividing by zero and ๐‘“(๐‘ฅ) would be undefined.

Let us consider what happens when we multiply two rational functions together. Recall that if we had two ordinary rational numbers (or fractions) given by ๐‘๐‘ž and ๐‘Ÿ๐‘ , their product is simply ๐‘๐‘žร—๐‘Ÿ๐‘ =๐‘๐‘Ÿ๐‘ž๐‘ .

That is to say we multiply the numerators (i.e., the tops) together and the denominators (i.e., the bottoms) together. Rational functions work in exactly the same way. Suppose we have two rational functions ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ) and ๐‘”(๐‘ฅ)=๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ). Then, their product is ๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)ร—๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ)=๐‘(๐‘ฅ)๐‘Ÿ(๐‘ฅ)๐‘ž(๐‘ฅ)๐‘ (๐‘ฅ).

The only difference is that we are now dealing with functions that depend on ๐‘ฅ rather than just numbers. Naturally, we still need to consider what domain is appropriate for this function. Using what we know about rational functions, we can say that ๐‘(๐‘ฅ)๐‘Ÿ(๐‘ฅ)๐‘ž(๐‘ฅ)๐‘ (๐‘ฅ) is only valid when the denominator ๐‘ž(๐‘ฅ)๐‘ (๐‘ฅ)โ‰ 0, which means we need both ๐‘ž(๐‘ฅ)โ‰ 0 and ๐‘ (๐‘ฅ)โ‰ 0 (since if either of them were zero, their product would be zero too). This leads to the following rule.

Rule: Product of Rational Functions

Let ๐‘”(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ) and โ„Ž(๐‘ฅ)=๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ) be two rational functions, and suppose their product is ๐‘“(๐‘ฅ)=๐‘”(๐‘ฅ)โ„Ž(๐‘ฅ). Then, ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘Ÿ(๐‘ฅ)๐‘ž(๐‘ฅ)๐‘ (๐‘ฅ) and the domain of ๐‘“(๐‘ฅ) is the common domain of ๐‘”(๐‘ฅ) and โ„Ž(๐‘ฅ).

Recall that the common domain of two rational functions is just the intersection of their domains, and we can calculate this by working out all the points ๐‘ฅ that cause either of the denominators ๐‘ž(๐‘ฅ) or ๐‘ (๐‘ฅ) to be zero and subtracting those points from โ„.

Let us demonstrate how this works with a simple example. Suppose we have ๐‘”(๐‘ฅ)=2๐‘ฅโˆ’3,โ„Ž(๐‘ฅ)=4๐‘ฅ+1๐‘ฅ.

To work out their product, ๐‘“(๐‘ฅ), we simply multiply the numerators and the denominators together. ๐‘“(๐‘ฅ)=2๐‘ฅโˆ’3ร—4๐‘ฅ+1๐‘ฅ=2(4๐‘ฅ+1)๐‘ฅ(๐‘ฅโˆ’3).

To work out the domain of this function, we find the common domain of ๐‘”(๐‘ฅ) and โ„Ž(๐‘ฅ). The domain of ๐‘”(๐‘ฅ) is โ„โˆ’{3}, and the domain of โ„Ž(๐‘ฅ) is โ„โˆ’{0}. Therefore, the domain of ๐‘“(๐‘ฅ) is the intersection of these two domains, โ„โˆ’{0,3}.

We can verify this by looking at the denominator of ๐‘“(๐‘ฅ). We can see that if we took ๐‘ฅ=0 or ๐‘ฅ=3, the denominator would be 0, meaning the expression would be undefined. Otherwise, any other value for ๐‘ฅโˆˆโ„ would be valid.

It is important to note that we always need to check that the domain is valid before simplifying by canceling any terms. Suppose, for example, we had ๐‘“(๐‘ฅ)=๐‘ฅ๐‘ฅโˆ’1ร—๐‘ฅโˆ’1๐‘ฅ.๏Šจ

If we simplify this function before checking the domain, we have ๐‘ฅ๐‘ฅ+1ร—๐‘ฅโˆ’1๐‘ฅ=๐‘ฅ(๐‘ฅโˆ’1)๐‘ฅ(๐‘ฅโˆ’1)=๐‘ฅ๐‘ฅ(๐‘ฅโˆ’1)๐‘ฅ(๐‘ฅโˆ’1)=๐‘ฅ.๏Šจ๏Šจ

At first, it looks like this function is valid for any ๐‘ฅโˆˆโ„. But this is discounting the fact that ๐‘“(๐‘ฅ) is undefined if we take ๐‘ฅ=0 or ๐‘ฅ=1 in the original expression. So, it is crucial that we check for any points where the denominator is 0 before canceling any terms.

Now, up to this point, we have only considered rational functions containing linear polynomials, but keep in mind that we will also need to consider rational functions that involve higher-order polynomials like quadratics as well. In such a situation, the best approach is usually to simplify the expressions via factoring as much as possible before multiplying them together.

Recall that, to factor a quadratic, we can apply the following general process.

How To: Factoring a Quadratic Expression

Let ๐‘“(๐‘ฅ)=๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘๏Šจ be a quadratic function. Then, we can apply a step-by-step process to factor ๐‘“(๐‘ฅ) as follows:

  1. Find ๐‘Ž๐‘ and what factors it has. So, look for numbers ๐‘ and ๐‘ž such that ๐‘๐‘ž=๐‘Ž๐‘.
  2. Find two factors of ๐‘Ž๐‘ that add together to get ๐‘. Let us say we find that ๐‘+๐‘ž=๐‘.
  3. Rewrite the expression as ๐‘Ž๐‘ฅ+๐‘๐‘ฅ+๐‘ž๐‘ฅ+๐‘๏Šจ.
  4. Factor the first two terms and the last two terms.
  5. Finally, factor out the common factor.

So, let us apply this to an example and say we have 2๐‘ฅ+7๐‘ฅ+6๏Šจ. Here, ๐‘Ž=2, ๐‘=7, and ๐‘=6. Applying the above process, we have

  1. ๐‘Ž๐‘=12. Its factors are (1,12), (2,6), and (3,4).
  2. Adding these factors together, ๏ฒ1+12=13,2+6=8,3+4=7. Since ๐‘=7, this means 3 and 4 are values that work.
  3. We rewrite the expression as 2๐‘ฅ+3๐‘ฅ+4๐‘ฅ+6๏Šจ.
  4. We factor out ๐‘ฅ from the first two terms and 2 from the second two terms to get ๐‘ฅ(2๐‘ฅ+3)+2(2๐‘ฅ+3).
  5. We factor (2๐‘ฅ+3) out as a common factor to get (๐‘ฅ+2)(2๐‘ฅ+3).

Finally, we can verify that this factoring is correct by multiplying out the parentheses, which shows that this is indeed equal to 2๐‘ฅ+7๐‘ฅ+6๏Šจ. In addition to this method, we also note that sometimes, we can apply shortcuts, such as noting ๐‘ฅโˆ’๐‘=(๐‘ฅ+๐‘)(๐‘ฅโˆ’๐‘)๏Šจ๏Šจ, which saves us a bit of working.

So, let us consider a product of rational expressions involving quadratic terms and see the general process that we use.

Example 1: Simplifying a Rational Function Involving the Product of Two Rational Expressions and Finding the Domain of the Resulting Function

Simplify the function ๐‘›(๐‘ฅ)=๐‘ฅ+16๐‘ฅ+64๐‘ฅ+8๐‘ฅร—7๐‘ฅโˆ’5664โˆ’๐‘ฅ๏Šจ๏Šจ๏Šจ, and determine its domain.


The best way to approach a question like this is to begin by simplifying the expression by factoring it where possible. Looking at the components of this expression one-by-one, we note that the numerator of the first rational function is a perfect square quadratic and can thus be factored by ๐‘ฅ+16๐‘ฅ+64=(๐‘ฅ+8).๏Šจ๏Šจ

In the denominator, we can factor out ๐‘ฅ to find that ๐‘ฅ+8๐‘ฅ=๐‘ฅ(๐‘ฅ+8).๏Šจ

In the second rational function, we can factor out 7 to get 7๐‘ฅโˆ’56=7(๐‘ฅโˆ’8).

Finally, in the second denominator, we have a quadratic that is the difference of two squares and can be factored by 64โˆ’๐‘ฅ=(8+๐‘ฅ)(8โˆ’๐‘ฅ).๏Šจ

Having factored all the terms, we can rewrite ๐‘›(๐‘ฅ) as follows: ๐‘›(๐‘ฅ)=(๐‘ฅ+8)๐‘ฅ(๐‘ฅ+8)ร—7(๐‘ฅโˆ’8)(8+๐‘ฅ)(8โˆ’๐‘ฅ).๏Šจ

Before we multiply these expressions together, it is important that we note that, for the first expression, the points ๐‘ฅ=0 and ๐‘ฅ=โˆ’8 are invalid, and for the second, the points ๐‘ฅ=โˆ’8 and ๐‘ฅ=8 are invalid. Using the fact that the domain of the product of rational functions is the common domain of those functions, we therefore know that the domain of ๐‘›(๐‘ฅ) must be โ„โˆ’{โˆ’8,0,8}.

We note that it is important to do this before simplifying further, because even if we cancel out terms, we are still not allowed to let ๐‘ฅ have values that are invalid in the original expression.

Now, before multiplying the two expressions together, we note that we can already cancel out terms in the numerators and the denominators. Noting that (8โˆ’๐‘ฅ)=โˆ’(๐‘ฅโˆ’8) in the denominator and that we can take the โˆ’ term outside the fraction, we have ๐‘›(๐‘ฅ)=(๐‘ฅ+8)๐‘ฅ(๐‘ฅ+8)ร—๏€ฝโˆ’7(๐‘ฅโˆ’8)(8+๐‘ฅ)(๐‘ฅโˆ’8)๏‰=(๐‘ฅ+8)๐‘ฅ(๐‘ฅ+8)ร—๏€พโˆ’7(๐‘ฅโˆ’8)(8+๐‘ฅ)(๐‘ฅโˆ’8)๏Š=๐‘ฅ+8๐‘ฅร—๏€ผโˆ’78+๐‘ฅ๏ˆ.๏Šจ๏Šจ

Now, we multiply the fractions together as follows: ๐‘›(๐‘ฅ)=๐‘ฅ+8๐‘ฅร—๏€ผโˆ’78+๐‘ฅ๏ˆ=โˆ’7(๐‘ฅ+8)๐‘ฅ(8+๐‘ฅ)=โˆ’7(๐‘ฅ+8)๐‘ฅ(8+๐‘ฅ)=โˆ’7๐‘ฅ.

So, in conclusion, we find that ๐‘›(๐‘ฅ)=โˆ’7๐‘ฅ and its domain is โ„โˆ’{โˆ’8,0,8}.

Let us consider a similar example, this time with a cubic expression in the numerator, and apply the same concepts we have just used.

Example 2: Simplifying a Rational Function Involving the Product of Two Rational Expressions and Finding the Domain of the Resulting Function

Simplify the function ๐‘›(๐‘ฅ)=๐‘ฅ+3432๐‘ฅ+14๐‘ฅร—๐‘ฅ+3๐‘ฅโˆ’7๐‘ฅ+49๏Šฉ๏Šจ๏Šจ, and determine its domain.


Let us start by factoring the two expressions as much as possible. Going through the numerators and denominators one by one, we start with ๐‘ฅ+343.๏Šฉ

This is a cubic expression. Recall that, for factoring the sum of cubes, we have the following formula we can use: ๐‘ฅ+๐‘=(๐‘ฅ+๐‘)๏€น๐‘ฅโˆ’๐‘๐‘ฅ+๐‘๏….๏Šฉ๏Šฉ๏Šจ๏Šจ

Let us determine if our function is of the correct form (i.e., is there a ๐‘ such that ๐‘=343๏Šฉ?). Going through 2,3,4โ€ฆ๏Šฉ๏Šฉ๏Šฉ, we see that 7=343๏Šฉ. This means that ๐‘=7 and we can apply the formula to obtain ๐‘ฅ+343=(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏….๏Šฉ๏Šจ

We note that, as ๐‘ฅ+343=0๏Šฉ is a sum of cubes, it only has one real solution, ๐‘ฅ=โˆ’7. Hence, the other factor of the expression, ๐‘ฅโˆ’7๐‘ฅ+49=0๏Šจ, has no real solutions and cannot be factored further.

Let us consider the other terms in ๐‘›(๐‘ฅ). The denominator of the first term can be factored by taking out 2๐‘ฅ: 2๐‘ฅ+14๐‘ฅ=2๐‘ฅ(๐‘ฅ+7).๏Šจ

In the numerator of the second term, we have ๐‘ฅ+3, which is already in its simplest form, and in the denominator, we once again have ๐‘ฅโˆ’7๐‘ฅ+49๏Šจ, which we know has no real solutions. All in all, this gives us ๐‘›(๐‘ฅ)=(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏…2๐‘ฅ(๐‘ฅ+7)ร—๐‘ฅ+3๐‘ฅโˆ’7๐‘ฅ+49.๏Šจ๏Šจ

Before simplifying further, let us find what values ๐‘ฅ is valid for in the domain. In the first fraction, we have 2๐‘ฅ(๐‘ฅ+7)โ‰ 0, which means ๐‘ฅโ‰ 0 and ๐‘ฅโ‰ โˆ’7. For the second, we know that ๐‘ฅโˆ’7๐‘ฅ+49=0๏Šจ has no real solutions, so any value for ๐‘ฅ is fine. Therefore, the domain of ๐‘›(๐‘ฅ) is โ„โˆ’{โˆ’7,0}.

Let us now multiply ๐‘›(๐‘ฅ) and simplify where possible: ๐‘›(๐‘ฅ)=(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏…2๐‘ฅ(๐‘ฅ+7)ร—๐‘ฅ+3๐‘ฅโˆ’7๐‘ฅ+49=(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏…(๐‘ฅ+3)2๐‘ฅ(๐‘ฅ+7)(๐‘ฅโˆ’7๐‘ฅ+49)=(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏…(๐‘ฅ+3)2๐‘ฅ(๐‘ฅ+7)๏€น๐‘ฅโˆ’7๐‘ฅ+49๏…=๐‘ฅ+32๐‘ฅ.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

So, in the end, we find that ๐‘›(๐‘ฅ)=(๐‘ฅ+3)2๐‘ฅ, with domain โ„โˆ’{โˆ’7,0}.

While we often have to simplify and find the domain of the product of rational expressions, sometimes we are just asked to find the value of that product at a certain point. In this situation, we simply have to substitute the given value into the function. It is important to pay attention to what the question is asking so that we can avoid doing work that is actually not necessary.

Example 3: Simplifying a Rational Function Involving the Product of Two Rational Expressions and Evaluating the Resulting Function at a Given Value

Given the function ๐‘›(๐‘ฅ)=๐‘ฅโˆ’6๐‘ฅโˆ’15๐‘ฅ+54ร—๐‘ฅโˆ’3๐‘ฅโˆ’282๐‘ฅโˆ’15๐‘ฅ+7๏Šจ๏Šจ๏Šจ, evaluate ๐‘›(7), if possible.


Normally, for questions where we have to multiply two rational functions together, it is easiest to factor the equations first so that we can find their domains and cancel any common factors. In this situation, though, we only need to evaluate ๐‘› at a single point, so we can just substitute ๐‘ฅ=7 into ๐‘› and see what we get. For completeness, however, we will demonstrate what happens when you factor the equation. After some calculations, we will get ๐‘›(๐‘ฅ)=๐‘ฅโˆ’6(๐‘ฅโˆ’6)(๐‘ฅโˆ’9)ร—(๐‘ฅโˆ’7)(๐‘ฅ+4)(2๐‘ฅโˆ’1)(๐‘ฅโˆ’7).

We note here that, due to the factors in the denominators, the domain of ๐‘›(๐‘ฅ) is โ„โˆ’{0.5,6,7,9}. Since ๐‘ฅ=7 is not in the domain, this shows that evaluating ๐‘›(7) would result in an undefined quantity. If we were to continue further and cancel out the factors, we would get =๐‘ฅโˆ’6(๐‘ฅโˆ’6)(๐‘ฅโˆ’9)ร—(๐‘ฅโˆ’7)(๐‘ฅ+4)(2๐‘ฅโˆ’1)(๐‘ฅโˆ’7)=1๐‘ฅโˆ’9ร—๐‘ฅ+42๐‘ฅโˆ’1=๐‘ฅ+4(๐‘ฅโˆ’9)(2๐‘ฅโˆ’1).

Now, if we had not checked the domain earlier and tried to evaluate ๐‘›(๐‘ฅ) at ๐‘ฅ=7, we would have found 7+4(7โˆ’9)(7ร—2โˆ’1)=11(โˆ’2)(13)=โˆ’1126.

Even though we get a well-defined answer here, we are not actually evaluating ๐‘› at 7 but rather a modified version of ๐‘› with the ๐‘ฅ=7 singularity cut out. So, the answer is not valid. As discussed earlier, the easiest approach to this type of question is simply to substitute ๐‘ฅ=7 into the original equation, which gives us ๐‘›(7)=7โˆ’67โˆ’15ร—7+54ร—7โˆ’3ร—7โˆ’282ร—7โˆ’15ร—7+7=149โˆ’105+54ร—49โˆ’21โˆ’282ร—49โˆ’105+7=1โˆ’2ร—00=00.๏Šจ๏Šจ๏Šจ

As we already knew, the result is undefined since it is 00, meaning we cannot evaluate the function. Therefore, ๐‘›(7) is undefined.

We have now seen what happens if we multiply two rational fractions together, but what happens if we divide one rational fraction by another? To consider this, let us return to what happens if we are working with rational numbers: ๐‘๐‘žรท๐‘Ÿ๐‘ =๐‘๐‘žร—๐‘ ๐‘Ÿ=๐‘๐‘ ๐‘ž๐‘Ÿ.

The answer, as we can see, is to take the reciprocal of the divisor. Recall that the reciprocal of a number ๐‘ฅ is just 1๐‘ฅ. With a fraction, taking the reciprocal is the same as swapping the numerator and the denominator. So, the reciprocal of ๐‘Ÿ๐‘  is just ๐‘ ๐‘Ÿ. We can do the same thing with rational fractions.

Supposing we have two rational functions ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ) and ๐‘”(๐‘ฅ)=๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ), their quotient is ๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)รท๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)ร—๐‘ (๐‘ฅ)๐‘Ÿ(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ (๐‘ฅ)๐‘ž(๐‘ฅ)๐‘Ÿ(๐‘ฅ).

Once we take the reciprocal of the function, the rest of the calculation goes the same way as for multiplying functions together.

The only other thing we need to consider is the domain of the quotient, where we need to be careful. Note that, in the above expression, we first of all need ๐‘ž(๐‘ฅ) and ๐‘ (๐‘ฅ)โ‰ 0 for any ๐‘ฅ. Once we have taken the reciprocal, we can see that we also need ๐‘Ÿ(๐‘ฅ)โ‰ 0. So, compared to the multiplication case, we need to check an extra term in order to determine the correct domain. This leads to the following rule.

Rule: Quotient of Rational Functions

Recall that the zeros of a function ๐‘“(๐‘ฅ) are the points ๐‘ง where ๐‘“(๐‘ง)=0. If ๐‘”(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ) and โ„Ž(๐‘ฅ)=๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ) are two rational functions and their quotient is ๐‘“(๐‘ฅ)=๐‘”(๐‘ฅ)โ„Ž(๐‘ฅ), then ๐‘“(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ (๐‘ฅ)๐‘ž(๐‘ฅ)๐‘Ÿ(๐‘ฅ) and the domain of ๐‘“(๐‘ฅ) is โ„โˆ’๐‘(๐‘ž(๐‘ฅ))โˆ’๐‘(๐‘Ÿ(๐‘ฅ))โˆ’๐‘(๐‘ (๐‘ฅ)), where ๐‘ denotes the set of zeros of a function.

This rule means we need to consider all the points where a denominator might be 0 and exclude them from the domain.

We will demonstrate exactly how this works in the following example.

Example 4: Identifying the Domain of the Quotient of Two Rational Expressions

Determine the domain of the function ๐‘›(๐‘ฅ)=3๐‘ฅโˆ’15๐‘ฅโˆ’6รท6๐‘ฅโˆ’304๐‘ฅโˆ’24.


Recall that, for the domain of the quotient of two rational expressions, we need to calculate the points where ๐‘ฅโˆ’6=0,6๐‘ฅโˆ’30=0,4๐‘ฅโˆ’24=0.

We do not need to consider the equation 3๐‘ฅโˆ’15=0 since it will not cause problems when dividing. By looking at the given function, we can confirm that if ๐‘ฅโˆ’6=0 or 4๐‘ฅโˆ’24=0, ๐‘›(๐‘ฅ) will be undefined. This means that we cannot have ๐‘ฅ=6, since this leads to both expressions being 0.

We know that when we divide by a rational expression, that is the same as multiplying by the reciprocal. So, another way to write ๐‘›(๐‘ฅ) is ๐‘›(๐‘ฅ)=3๐‘ฅโˆ’15๐‘ฅโˆ’6รท6๐‘ฅโˆ’304๐‘ฅโˆ’24=3๐‘ฅโˆ’15๐‘ฅโˆ’6ร—4๐‘ฅโˆ’246๐‘ฅโˆ’30.

We note that, in order to find the domain, it is important not to simplify the expression further since we might cancel terms that are important. In this example, we have ๐‘ฅโˆ’6 in the denominator and 4๐‘ฅโˆ’24=4(๐‘ฅโˆ’6) in the numerator, leading to ๐‘ฅโˆ’6 canceling out. If we performed this before checking the domain, then we might miss the fact that ๐‘ฅโ‰ 6 (although, fortunately, we have already found this above).

In any case, by considering this form for ๐‘›(๐‘ฅ), we can confirm that 6๐‘ฅโˆ’30โ‰ 0 is the other requirement for our domain, which leads to ๐‘ฅโ‰ 5. Now knowing that the set of invalid points for this function is {5,6}, we can say that the domain must be โ„โˆ’{5,6}.

Having looked at the division of rational functions with linear components, let us once again return to the case of a problem involving quadratics, since this will result in us having to do slightly more in-depth calculations.

Example 5: Evaluating the Domain of the Quotient of Two Rational Expressions with Quadratic Components

Determine the domain of the function ๐‘›(๐‘ฅ)=๐‘ฅโˆ’๐‘ฅโˆ’6๐‘ฅโˆ’4รท2๐‘ฅโˆ’6๐‘ฅโˆ’4๐‘ฅ+4๏Šจ๏Šจ๏Šจ.


Recall that, for the domain of the quotient of two rational expressions, we need to consider the points where ๐‘ฅโˆ’4=0,2๐‘ฅโˆ’6=0,๐‘ฅโˆ’4๐‘ฅ+4=0.๏Šจ๏Šจ

We do not need to consider ๐‘ฅโˆ’๐‘ฅโˆ’6๏Šจ since it is not a problem if it is equal to 0. So, let us begin by factoring these three expressions. We can factor the first expression to get ๐‘ฅโˆ’4=0(๐‘ฅโˆ’2)(๐‘ฅ+2)=0.๏Šจ

So, here we find ๐‘ฅ=โˆ’2,2 are two points that are not valid in the domain. For our second expression, 2๐‘ฅโˆ’6=02(๐‘ฅโˆ’3)=0.

So, ๐‘ฅ=3 is another invalid point. Finally, we have ๐‘ฅโˆ’4๐‘ฅ+4=0(๐‘ฅโˆ’2)=0.๏Šจ๏Šจ

This tells us ๐‘ฅ=2 is invalid, which we already knew from the first expression.

Combining this together, we find that the domain of ๐‘›(๐‘ฅ) is โ„โˆ’{โˆ’2,2,3}.

As of yet, we have not had to calculate an explicit expression for the quotient of two rational functions. For our final example, let us consider a problem where we must do this and then find the point at which the resulting function is equal to a given value.

Example 6: Simplifying a Rational Function Using Factorization and Then Evaluating Its Variable at Given Values

Given that ๐‘“(๐‘ฅ)=๐‘ฅ+9๐‘ฅ+14๐‘ฅโˆ’4รท๐‘ฅโˆ’49๐‘ฅโˆ’2๐‘ฅ๏Šจ๏Šจ๏Šจ๏Šจ and ๐‘“(๐‘Ž)=4, find the value of ๐‘Ž.


To solve this problem, we could try substituting ๐‘Ž into ๐‘“ directly and setting it equal to 4, but as this would result in an equation in terms of ๐‘Ž that would not simplify easily, this would probably prove difficult. Instead, a better approach is to begin by simplifying the function and then substituting ๐‘Ž into the simplified form. Let us begin the simplification by starting with the top left numerator, ๐‘ฅ+9๐‘ฅ+14.๏Šจ

To factor this, since it is a quadratic expression, we want to find two numbers that multiply together to get 14 and add together to get 9. These are 2 and 7. Thus, we can factor the equation to get ๐‘ฅ+9๐‘ฅ+14=๐‘ฅ+2๐‘ฅ+7๐‘ฅ+14=(๐‘ฅ+2)(๐‘ฅ+7).๏Šจ๏Šจ

For the denominator, we notice that ๐‘ฅโˆ’4๏Šจ is a perfect square. Thus, we can factor it to find ๐‘ฅโˆ’4=(๐‘ฅ+2)(๐‘ฅโˆ’2).๏Šจ

For the second fraction, we can see that, in the numerator, we again have a perfect square that can be factored to get ๐‘ฅโˆ’49=(๐‘ฅ+7)(๐‘ฅโˆ’7).๏Šจ

Finally, for the denominator, we notice there is a common factor of ๐‘ฅ, which we can factor out to get ๐‘ฅโˆ’2๐‘ฅ=๐‘ฅ(๐‘ฅโˆ’2).๏Šจ

Putting this all together, we get the following factoring of ๐‘“(๐‘ฅ): ๐‘“(๐‘ฅ)=(๐‘ฅ+2)(๐‘ฅ+7)(๐‘ฅ+2)(๐‘ฅโˆ’2)รท(๐‘ฅ+7)(๐‘ฅโˆ’7)๐‘ฅ(๐‘ฅโˆ’2).

Now, we have been given that ๐‘“(๐‘Ž)=4 is a valid equation, but just to be safe, we can find the domain of ๐‘“ before simplifying further, by calculating the zeros of the following equations: (๐‘ฅ+2)(๐‘ฅโˆ’2)=0,(๐‘ฅ+7)(๐‘ฅโˆ’7)=0,๐‘ฅ(๐‘ฅโˆ’2)=0.

The first equation gives us ๐‘ฅ=โˆ’2 and ๐‘ฅ=2, the second gives us ๐‘ฅ=โˆ’7 and ๐‘ฅ=7, and the third gives ๐‘ฅ=0 and ๐‘ฅ=2 again. Thus, the domain of ๐‘“(๐‘ฅ) is โ„โˆ’{โˆ’7,โˆ’2,0,2,7}.

Now, we simplify the equation for ๐‘“ by taking the reciprocal of the divisor to get ๐‘“(๐‘ฅ)=(๐‘ฅ+2)(๐‘ฅ+7)(๐‘ฅ+2)(๐‘ฅโˆ’2)ร—๐‘ฅ(๐‘ฅโˆ’2)(๐‘ฅ+7)(๐‘ฅโˆ’7).

Next, we cancel out the factors as follows: ๐‘“(๐‘ฅ)=(๐‘ฅ+2)(๐‘ฅ+7)(๐‘ฅ+2)(๐‘ฅโˆ’2)ร—๐‘ฅ(๐‘ฅโˆ’2)(๐‘ฅ+7)(๐‘ฅโˆ’7)=๐‘ฅ๐‘ฅโˆ’7.

Thanks to the simplifications, this equation has now been significantly reduced to the point where we can solve ๐‘“(๐‘Ž)=4 quite easily. We have ๐‘Ž๐‘Žโˆ’7=4๐‘Ž=4(๐‘Žโˆ’7)๐‘Ž=4๐‘Žโˆ’28โˆ’3๐‘Ž=โˆ’28๐‘Ž=283.

Let us finish by going over the key points we have learned during this explainer.

Key Points

  • Multiplying rational functions together works as follows: ๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)ร—๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ)=๐‘(๐‘ฅ)๐‘Ÿ(๐‘ฅ)๐‘ž(๐‘ฅ)๐‘ (๐‘ฅ), where the domain of the product is โ„โˆ’๐‘(๐‘ž)โˆ’๐‘(๐‘ ) and ๐‘ denotes the zeros of a function.
  • Dividing rational functions works in almost the same way: ๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)รท๐‘Ÿ(๐‘ฅ)๐‘ (๐‘ฅ)=๐‘(๐‘ฅ)๐‘ž(๐‘ฅ)ร—๐‘ (๐‘ฅ)๐‘Ÿ(๐‘ฅ)=๐‘(๐‘ฅ)๐‘ (๐‘ฅ)๐‘ž(๐‘ฅ)๐‘Ÿ(๐‘ฅ), where the domain of the quotient is โ„โˆ’๐‘(๐‘ž)โˆ’๐‘(๐‘ )โˆ’๐‘(๐‘Ÿ).
  • Factoring helps us to simplify the answer and find any points in the domain that are not valid.
  • When calculating the quotient, we must be careful to check the numerator of the divisor when identifying the domain.

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