### 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.