# Lesson Video: Multiplying and Dividing Rational Functions Mathematics

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

22:01

### Video Transcript

In this video, we will learn how to multiply and divide rational functions. A function 𝑓 mapping a domain set 𝑋 to arrange set 𝑌 is called a rational function if it can be written in the form 𝑓 of 𝑥 equals 𝑝 of 𝑥 over 𝑞 of 𝑥, where 𝑝 and 𝑞 are polynomial functions and 𝑞 of 𝑥 is not equal to zero for all values 𝑥 in the domain set 𝑋. The domain set 𝑋 depends on the denominator function 𝑞. Any 𝑥-values for which 𝑞 of 𝑥 is equal to zero must be excluded from the domain set 𝑋 because otherwise we would be dividing by zero and 𝑓 of 𝑥 would be undefined.

Let’s consider what happens when we multiply two rational functions together. Recall that if we take two rational numbers, 𝑝 over 𝑞 and 𝑟 over 𝑠, and multiply them together, their product is 𝑝𝑟 over 𝑞𝑠. We multiply the numerators 𝑝 and 𝑟 together to give the new numerator 𝑝𝑟. And we multiply the denominators 𝑞 and 𝑠 together to give the new denominator 𝑞𝑠. Rational functions work in exactly the same way. Suppose we have two rational functions: 𝑔 of 𝑥 equals 𝑝 of 𝑥 over 𝑞 of 𝑥 and ℎ of 𝑥 equals 𝑟 of 𝑥 over 𝑠 of 𝑥. To take the product of these two functions, we treat them the same way we would rational numbers. We take the numerators and multiply them together to give the new numerator. And we take the denominators and multiply them together to give the new denominator.

This leads to the following result. Let 𝑔 of 𝑥 equal 𝑝 of 𝑥 over 𝑞 of 𝑥 and ℎ of 𝑥 equal 𝑟 of 𝑥 over 𝑠 of 𝑥 be two rational functions. And suppose their product 𝑔 of 𝑥 ℎ of 𝑥 is equal to 𝑓 of 𝑥. Then 𝑓 of 𝑥 equals 𝑝 of 𝑥 𝑟 of 𝑥 over 𝑞 of 𝑥 𝑠 of 𝑥, and the domain of 𝑓 of 𝑥 is the common domain of 𝑔 of 𝑥 and ℎ of 𝑥. The common domain of 𝑔 of 𝑥 and ℎ of 𝑥 is the domain of 𝑔 intersected with the domain of ℎ, that is, all of the elements which are common to both domains. We can find this common domain by finding all the values of 𝑥 which cause either 𝑞 of 𝑥 or 𝑠 of 𝑥 to be zero and therefore 𝑓 of 𝑥 to be undefined and removing them from the set of real numbers.

Let’s consider a simple example. 𝑔 of 𝑥 equals two over 𝑥 minus three, and ℎ of 𝑥 equals four 𝑥 plus one over 𝑥. To take the product of 𝑔 of 𝑥 and ℎ of 𝑥, we take the numerators and multiply them together and we take the denominators and multiply them together. This gives two times four 𝑥 plus one over 𝑥 minus three times 𝑥. The domain of 𝑓 is the real numbers minus all of the values of 𝑥 which cause 𝑓 of 𝑥 to be undefined. The values of 𝑥 which cause 𝑓 of 𝑥 to be undefined are the same values of 𝑥 that cause 𝑔 of 𝑥 and ℎ of 𝑥 to be undefined.

The value of 𝑥 that causes 𝑔 of 𝑥 to be undefined satisfies the equation 𝑥 minus three equals zero. Therefore, 𝑔 of 𝑥 is undefined at 𝑥 equals three. The value of 𝑥 which causes ℎ of 𝑥 to be undefined satisfies the equation 𝑥 equals zero. Therefore, ℎ of 𝑥 is undefined at 𝑥 equals zero. Since 𝑓 of 𝑥 is the product of 𝑔 of 𝑥 and ℎ of 𝑥, it is undefined at both 𝑥 equals zero and 𝑥 equals three. Therefore, the domain of 𝑓 is the set of real numbers ℝ subtract the set of values zero and three. When taking the product of two rational functions, we always need to find the domain of the resultant function before simplifying it.

Consider, for example, if we had 𝑓 of 𝑥 equals 𝑥 squared over 𝑥 minus one times 𝑥 minus one over 𝑥. Taking this product gives us 𝑥 squared times 𝑥 minus one over 𝑥 minus one times 𝑥. If we attempt to simplify this expression first by canceling the 𝑥 minus ones on the numerator and denominator and then one of the 𝑥’s on the numerator with the 𝑥 on the denominator, this would give us 𝑓 of 𝑥 equals 𝑥, a function which is defined for any real value of 𝑥. But this ignores the fact that the original product was undefined at two values of 𝑥, 𝑥 equals one and 𝑥 equals zero. It’s therefore crucial that we verify the domain of 𝑓 before canceling any terms in the expression.

Now let’s look at an example question.

Simplify the function 𝑓 of 𝑥 equals 𝑥 squared plus 16𝑥 plus 64 over 𝑥 squared plus eight 𝑥 times seven 𝑥 minus 56 over 64 minus 𝑥 squared, and determine its domain.

The best way to approach a problem like this is to simplify the expressions by factorization before taking the product. In the first quotient, both the numerator and the denominator are quadratics and can be factorized. Let’s consider the numerator first. This expression can be factorized into 𝑥 plus eight times 𝑥 plus eight, which can be rewritten as 𝑥 plus eight all squared. Now, let’s look at the denominator. This can be factorized into 𝑥 times 𝑥 plus eight. Now, let’s look at the numerator of the second term. Here, we can take out a common factor of seven to give seven times 𝑥 minus eight. And finally, for the denominator of the second term, we have a difference of two squares since 64 is equal to eight squared. We can therefore factorize this into eight minus 𝑥 times eight plus 𝑥.

We can therefore simplify the whole of 𝑓 of 𝑥 into 𝑥 plus eight all squared over 𝑥 times 𝑥 plus eight times seven times 𝑥 minus eight over eight minus 𝑥 times eight plus 𝑥. We now need to find the domain of 𝑓 of 𝑥 before taking the product and simplifying with cancelations. 𝑓 of 𝑥 is undefined if any of the terms in the products on the denominator are equal to zero. So if 𝑥 is equal to zero, 𝑥 plus eight is equal to zero, eight minus 𝑥 is equal to zero, or eight plus 𝑥 is equal to zero.

The values of 𝑥 for which 𝑓 of 𝑥 is undefined are the values which satisfy these equations. So we have 𝑥 equals zero, 𝑥 equals negative eight, 𝑥 equals eight, and 𝑥 equals negative eight once more. Therefore, the domain of 𝑓 is the set of real numbers ℝ subtract the set of values negative eight, zero, and eight. And notice that we do not need to include negative eight twice.

Now, we can proceed with canceling terms to simplify the expressions. For the first term in the product, we can cancel one of the 𝑥 plus eights on the numerator with the 𝑥 plus eight on the denominator. And for the second term, notice that one of the terms on the numerator, 𝑥 minus eight, is exactly negative one times one of the terms on the denominator, eight minus 𝑥. If we take out a factor of negative one from the 𝑥 minus eight on the numerator, it becomes eight minus 𝑥. This will then cancel with the term on the denominator. So far then, we have simplified 𝑓 of 𝑥 to 𝑥 plus eight over 𝑥 times negative seven over eight plus 𝑥.

Now taking this product by multiplying the numerators together and the denominators together, we get 𝑥 plus eight times negative seven over 𝑥 times eight plus 𝑥. The 𝑥 plus eight on the numerator will cancel with the eight plus 𝑥 on the denominator. This gives us the second part of our answer, 𝑓 of 𝑥 equals negative seven over 𝑥.

In this example, we were dealing with quadratics on the numerators and denominators. Let’s look at an example where we have cubic expressions.

Simplify the function 𝑓 of 𝑥 equals 𝑥 cubed plus 343 over two 𝑥 squared plus 14𝑥 times 𝑥 plus three over 𝑥 squared minus seven 𝑥 plus 49, and determine its domain.

Before taking the product, it’s best to simplify the expressions as much as possible by factorization, since this will allow us to easily determine 𝑓 of 𝑥’s domain. Let’s start with the numerator of the first term, 𝑥 cubed plus 343. Recall that if we have a sum of two cubes, 𝑥 cubed and 𝑐 cubed, we can factorize this expression as 𝑥 plus 𝑐 times 𝑥 squared minus 𝑐𝑥 plus 𝑐 squared. The expression here is a sum of two cubes since we have 𝑥 cubed and 343, which will be something cubed. In fact, 343 happens to be precisely seven cubed. Therefore, we can factorize this expression to give 𝑥 plus seven times 𝑥 squared minus seven 𝑥 plus seven squared, which is 49.

We could try factorizing this further by factorizing this quadratic term. However, if we consider the discriminant of this quadratic term, 𝑏 squared minus four 𝑎𝑐, we get negative 147, which is less than zero. Therefore, this expression has no real roots and therefore cannot be factorized. Next, looking at the denominator of the first term, two 𝑥 squared plus 14𝑥, this can be factorized by taking out a common factor of two 𝑥 to give two 𝑥 times 𝑥 plus seven. The numerator on the right-hand term, 𝑥 plus three, is already as simple as it can be. And for the denominator of the right-hand term, we have exactly the same expression as we had up here, 𝑥 squared minus seven 𝑥 plus 49. Therefore, this cannot be factorized any further.

We have so far simplified 𝑓 of 𝑥 to 𝑥 plus seven times 𝑥 squared minus seven 𝑥 plus 49 over two 𝑥 times 𝑥 plus seven timesed by 𝑥 plus three over 𝑥 squared minus seven 𝑥 plus 49. Before taking the product, we need to establish the domain of 𝑓 by finding all the values of 𝑥 for which 𝑓 of 𝑥 is undefined and taking them away from the set of real numbers. 𝑓 of 𝑥 will be undefined for any values of 𝑥 which cause either of the denominators in this product to be equal to zero. These values of 𝑥 will therefore satisfy two 𝑥 equals zero, 𝑥 plus seven equals zero, and 𝑥 squared minus seven 𝑥 plus 49 equals zero. These first two equations we can solve easily for 𝑥 to give 𝑥 equals zero and 𝑥 equals negative seven.

For the final equation, we have already shown that this quadratic expression has a discriminant of less than zero and therefore has no real solutions. Therefore, there are no values of 𝑥 which cause this denominator to be equal to zero. Therefore, the domain of 𝑓 is the set of real numbers ℝ subtract the values zero and seven. Now, we can proceed with taking the product of these two expressions by multiplying the numerators together and the denominators together to give 𝑥 plus seven times 𝑥 squared minus seven 𝑥 plus 49 times 𝑥 plus three all over two 𝑥 times 𝑥 plus seven times 𝑥 squared minus seven 𝑥 plus 49.

Since we’ve established the domain of 𝑓, we can now safely cancel terms. The 𝑥 plus seven on the numerator will cancel with the 𝑥 plus seven on the denominator. And the 𝑥 squared minus seven 𝑥 plus 49 on the numerator will cancel with the same on the denominator. And this gives us the second part of our answer, 𝑓 of 𝑥 equals 𝑥 plus three over two 𝑥.

Sometimes, instead of finding the domain of a product of rational functions, we simply need to evaluate the function at a given point. In this situation, we don’t need to simplify or cancel any terms, and we simply substitute the value of 𝑥 into the expression. Let’s look at an example.

Given the function 𝑓 of 𝑥 equals 𝑥 minus six over 𝑥 squared minus 15𝑥 plus 54 times 𝑥 squared minus three 𝑥 minus 28 over two 𝑥 squared minus 15𝑥 plus seven, evaluate 𝑓 of seven if possible.

To evaluate 𝑓 of seven, all we need to do is substitute the value 𝑥 equals seven into the expression. Doing so gives us 𝑓 of seven equals seven minus six over seven squared minus 15 times seven plus 54 all multiplied by seven squared minus three times seven minus 28 over two times seven squared minus 15 times seven plus seven. Simplifying both terms in the expression gives us one divided by negative two times zero divided by zero, which is undefined. Therefore, we cannot evaluate 𝑓 of seven because 𝑓 is undefined at this point.

We have so far explored how to multiply two rational functions together. But what if we want to divide one rational function by another? Consider how we do this with rational numbers, say, 𝑝 over 𝑞 and 𝑟 over 𝑠. To take 𝑝 over 𝑞 divided by 𝑟 over 𝑠, we reciprocate the second rational number, 𝑟 over 𝑠, and then take the product. This then gives us 𝑝 over 𝑞 multiplied by 𝑠 over 𝑟. We then proceed with the multiplication as normal, giving 𝑝𝑠 over 𝑞𝑟. Recall also that when dividing rational numbers, we need to take even more care with zeros.

To begin with, both 𝑞 and 𝑠 must be nonzero for the rational numbers to be defined. But then, when taking the product, 𝑟 must also be nonzero since we’re dividing by it here. The same applies to rational functions. If 𝑔 of 𝑥 equal to 𝑝 of 𝑥 over 𝑞 of 𝑥 and ℎ of 𝑥 equal to 𝑟 of 𝑥 over 𝑠 of 𝑥 are two rational functions and their quotient is 𝑓 of 𝑥 equals 𝑔 of 𝑥 over ℎ of 𝑥, then 𝑓 of 𝑥 equals 𝑝 of 𝑥 𝑠 of 𝑥 over 𝑞 of 𝑥 𝑟 of 𝑥. And the domain of 𝑓 is the set of real numbers ℝ minus 𝑍 of 𝑞 of 𝑥 minus 𝑍 of 𝑟 of 𝑥 minus 𝑍 of 𝑠 of 𝑥. 𝑍 denotes the zeros of a function.

So, for instance, 𝑍 of 𝑞 of 𝑥 is the set of 𝑥-values that satisfy 𝑞 of 𝑥 equals zero. We need to subtract the zeros of 𝑞 of 𝑥, 𝑟 of 𝑥, and 𝑠 of 𝑥 from the domain of 𝑓 by the same reasoning that 𝑞, 𝑟, and 𝑠 when dividing rational numbers cannot be equal to zero.

Let’s see exactly how this works with an example.

Determine the domain of the function 𝑓 of 𝑥 equals three 𝑥 minus 15 over 𝑥 minus six divided by six 𝑥 minus 30 over four 𝑥 minus 24.

Recall that when we have a function 𝑓 of 𝑥 that is the quotient of two rational functions, 𝑝 of 𝑥 over 𝑞 of 𝑥 and 𝑟 of 𝑥 over 𝑠 of 𝑥, we need to ensure that we exclude any values of 𝑥 from the domain of 𝑓 that satisfy 𝑞 of 𝑥 equals zero, 𝑟 of 𝑥 equals zero, and 𝑠 of 𝑥 equals zero. 𝑞 of 𝑥 and 𝑠 of 𝑥 must never be equal to zero because then we would be dividing by zero and 𝑓 would be undefined. In addition, 𝑟 of 𝑥 must also be not equal to zero since when taking the quotient here, we will be taking the reciprocal of 𝑟 of 𝑥 over 𝑠 of 𝑥 and then multiplying the two fractions. Therefore, the domain of 𝑓 is the set of real numbers ℝ minus 𝑍 of 𝑞 of 𝑥 minus 𝑍 of 𝑟 of 𝑥 minus 𝑍 of 𝑠 of 𝑥, where 𝑍 denotes the zeros of the function, that is, the values of 𝑥 for which the function is equal to zero.

In our case, 𝑞 of 𝑥 is equal to 𝑥 minus six, 𝑟 of 𝑥 is six 𝑥 minus 30, and 𝑠 of 𝑥 is four 𝑥 minus 24. We therefore need to solve these three equations for 𝑥 and subtract these values of 𝑥 from the set of real numbers to give the domain of 𝑓. The first equation solves to 𝑥 equals six, the second equation solves to 𝑥 equals five, and the third equation solves to 𝑥 equals six once again. Therefore, the domain of 𝑓 is the set of real numbers ℝ minus the set of values five and six. And notice we do not have to subtract six twice.

In the final example, we will look at the quotient of rational functions containing cubic expressions.

Simplify the function 𝑓 of 𝑥 equals 𝑥 squared minus 12𝑥 plus 36 over 𝑥 cubed minus 216 divided by seven 𝑥 minus 42 over 𝑥 squared plus six 𝑥 plus 36, and determine its domain.

To simplify 𝑓 of 𝑥, let’s begin by factorizing each of the terms in the rational expressions, starting with the numerator of the left-hand term, 𝑥 squared minus 12𝑥 plus 36. This readily factorizes to 𝑥 minus six times 𝑥 minus six, which is 𝑥 minus six all squared. For the denominator of the left-hand term, we have a difference of two cubes. Recall that when we have a difference of two cubes, 𝑥 cubed minus 𝑐 cubed, we can factorize this as 𝑥 minus 𝑐 times 𝑥 squared plus 𝑐𝑥 plus 𝑐 squared. 216 will be something cubed, and it so happens that it’s a perfect cube, six cubed. Therefore, we can factorize this as 𝑥 minus six times 𝑥 squared plus six 𝑥 plus six squared, which is 36.

We could try to factorize this quadratic expression further. By taking the discriminant 𝑏 squared minus four 𝑎𝑐, we get a result of negative 108, which is less than zero. Therefore, this quadratic expression has no real roots and cannot be factorized further. For the numerator of the second rational function, we have seven 𝑥 minus 42. We can take out a common factor of seven to give seven times 𝑥 minus six. And finally, for the denominator of the second rational function, we have 𝑥 squared plus six 𝑥 plus 36. We have just shown that this exact expression has no real roots and cannot be factorized further.

So far then, we can simplify 𝑓 of 𝑥 to 𝑥 minus six all squared over 𝑥 minus six times 𝑥 squared plus six 𝑥 plus 36 divided by seven times 𝑥 minus six over 𝑥 squared plus six 𝑥 plus 36. Recall that if we have a function 𝑓 of 𝑥 that is the quotient of two rational functions, 𝑝 of 𝑥 over 𝑞 of 𝑥 and 𝑟 of 𝑥 over 𝑠 of 𝑥, the domain of 𝑓 is the set of real numbers ℝ minus 𝑍 of 𝑞 of 𝑥 minus 𝑍 of 𝑟 of 𝑥 minus 𝑍 of 𝑠 of 𝑥, where 𝑍 denotes the zeros of a function. In our case, 𝑞 of 𝑥 is the denominator of the first rational function, 𝑥 minus six times 𝑥 squared plus six 𝑥 plus 36. 𝑟 of 𝑥 is the numerator of the second rational function, seven times 𝑥 minus six. And 𝑠 of 𝑥 is the denominator of the second rational function, 𝑥 squared plus six 𝑥 plus 36.

We need to find all the values of 𝑥 that solve the equations 𝑞 of 𝑥 equals zero, 𝑟 of 𝑥 equals zero, and 𝑠 of 𝑥 equals zero and subtract them from the set of real numbers to find the domain of 𝑓. For the first equation, either 𝑥 minus six must equal zero or 𝑥 squared plus six 𝑥 plus 36 must equal zero. If 𝑥 minus six equals zero, 𝑥 is equal to six.

For this quadratic expression, we already saw earlier that the discriminant, 𝑏 squared minus four 𝑎𝑐, was less than zero. Therefore, this expression has no real roots and therefore cannot equal zero for any real value of 𝑥. The second equation solves easily to give 𝑥 equals six once again. And for the third equation, we have exactly the same quadratic expression with the discriminant less than zero. Therefore, there are no real values of 𝑥 that solve this equation. Therefore, the only value of 𝑥 for which any of these functions are equal to zero is six. Therefore, the domain of 𝑓 is the set of real numbers ℝ minus the single element six.

Now, we can safely proceed with taking the quotient and simplifying. To take the quotient, we reciprocate the dividing rational function. So we swap the numerator and denominator and then we change the division to multiplication. Taking the multiplication by multiplying the numerators together and the denominators together, we get 𝑥 minus six all squared times 𝑥 squared plus six 𝑥 plus 36 all over 𝑥 minus six times 𝑥 squared plus six 𝑥 plus 36 times seven times 𝑥 minus six. The 𝑥 minus six all squared on the numerator will cancel with both 𝑥 minus six terms on the denominator. And the 𝑥 squared plus six 𝑥 plus 36 will cancel with the same on the denominator. This leaves us with just one on the numerator and seven on the denominator, so 𝑓 of 𝑥 equals one-seventh.

Let’s finish this video by recapping some key points. If we have two rational functions, 𝑝 of 𝑥 over 𝑞 of 𝑥 and 𝑟 of 𝑥 over 𝑠 of 𝑥, their product is given by 𝑝 of 𝑥 𝑟 of 𝑥 over 𝑞 of 𝑥 𝑠 of 𝑥. The domain of the resulting function is the set of real numbers ℝ subtract any values of 𝑥 which cause the function to be undefined, that is, the zeros of 𝑞 of 𝑥 and 𝑠 of 𝑥. To take the quotient to two rational functions, we reciprocate the dividing rational function, 𝑟 of 𝑥 over 𝑠 of 𝑥, by switching the numerator with the denominator. And then we take the product of the two resulting rational functions, giving 𝑝 of 𝑥 𝑠 of 𝑥 over 𝑞 of 𝑥 𝑟 of 𝑥. The domain of the resulting function will be the set of real numbers ℝ subtract the zeros of 𝑞 of 𝑥, 𝑟 of 𝑥, and 𝑠 of 𝑥.

Remember that we need to subtract the zeros of 𝑟 of 𝑥, since we will be reciprocating it and then dividing by it. So it being equal to zero will also cause the function to be undefined. Remember that we always need to find the domain of the resultant function before simplifying and canceling any terms.

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