### Video Transcript

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

We begin by recalling that a
rational function is the quotient of two polynomials. The function π of π₯, which is
equal to π of π₯ over π of π₯, is a rational function if both π of π₯ and π of
π₯ are polynomials, where π of π₯ is not the zero polynomial. Whilst simplifying rational
functions follows a similar process to simplifying fractions, there is one added
complication. This involves the domain of the
function, and we will consider this first.

If π of π₯, which is equal to π
of π₯ over π of π₯, is a rational function, then the domain of π of π₯ is all real
numbers except those where π of π₯ is equal to zero. This is due to the fact that when
the denominator of a function is zero, the function is undefined. Letβs consider the function π of
π₯ is equal to π₯ plus four over π₯ minus six. This function will be undefined
when the denominator is equal to zero. So, we set π₯ minus six equals zero
and solve this equation to help find the domain of π of π₯. Adding six to both sides of our
equation, we have π₯ equals six. And we can therefore conclude that
the domain of π of π₯ is the set of all real values apart from six. This is because the function is
defined for all values of π₯ apart from six.

Before looking at some examples, we
will briefly look at a four-step process we can use to simplify rational
functions. To simplify the rational function
π of π₯ is equal to π of π₯ over π of π₯, we follow the following steps. Firstly, we find all the values of
π₯ where π of π₯ is equal to zero, as the domain of π of π₯ is all real values
except these roots. Our second step is to fully factor
both π of π₯ and π of π₯. Next, we cancel any shared factors
in the numerator and denominator, noting that we must still restrict the values of
π₯ to be in the domain of π of π₯. Finally, we equate π of π₯ to the
simplified expression restricted to the domain of π of π₯. We will now consider how we can
apply this process to an example.

Simplify the function π of π₯
is equal to π₯ squared plus two π₯ over π₯ squared minus four and find its
domain.

We begin by noting that the
function π of π₯ is the quotient of two polynomials and is therefore a rational
function. And in order to simplify a
rational function, we find its domain, factor the numerator and denominator, and
then cancel shared factors over the domain. We will therefore begin by
finding the domain of the function. In order to do this, we recall
that the domain of a rational function is all real values of π₯ except those
where the denominator is equal to zero.

Setting the denominator π of
π₯ equal to zero, we have π₯ squared minus four equals zero. The left-hand side can be
factored using the difference of two squares. π₯ squared minus four is equal
to π₯ plus two multiplied by π₯ minus two. For this to be equal to zero,
we know that one of the factors must equal zero: either π₯ plus two equals zero
or π₯ minus two equals zero. This gives us two solutions of
π₯ equals negative two and π₯ equals positive two. And it is for these two values
that the function π of π₯ is undefined. We can therefore conclude that
the domain of our function is the set of all real values minus the set
containing negative two and two.

As weβve already factored the
denominator, our next step is to factor the numerator. π₯ squared and two π₯ have a
common factor of π₯. So, we can rewrite this as π₯
multiplied by π₯ plus two. π of π₯ is therefore equal to
π₯ multiplied by π₯ plus two divided by π₯ plus two multiplied by π₯ minus
two. We can now cancel a common
factor of π₯ plus two from the numerator and denominator. We can do this since weβve
already established that π₯ equals negative two is not in the domain of π of
π₯. And the function simplifies to
π₯ over π₯ minus two.

We now have the two answers to
this question. The function π of π₯
simplifies to π₯ over π₯ minus two, and the domain of the function is the set of
real values minus the set containing negative two and two. In our next example, we will
use a simplification of a rational function to determine the value of an unknown
in the function.

Given that π of π₯ is equal to
π₯ squared plus 12π₯ plus 36 over π₯ squared minus π simplifies to π of π₯ is
equal to π₯ plus six over π₯ minus six, what is the value of π?

We begin by recalling that to
simplify a rational function, we find its domain, factor the numerator and
denominator, and then cancel the shared factors over the domain. In this question, the
simplified function has a linear denominator, whereas the original function has
a quadratic denominator. We must therefore be able to
factor π₯ squared minus π into two linear factors, and one of these must be π₯
minus six. We notice that our original
expression appears to be of the form π₯ squared minus π¦ squared. And this can be factored using
the difference of two squares into two linear factors: π₯ minus π¦, π₯ plus
π¦. Since π¦ is equal to six, we
have π₯ squared minus π is equal to π₯ minus six multiplied by π₯ plus six. And distributing the
parentheses or noting that π is equal to π¦ squared, the value of π is equal
to 36.

We can verify this by
considering the initial expression for π of π₯. If π is equal to 36, π of π₯
is equal to π₯ squared plus 12π₯ plus 36 over π₯ squared minus 36. The numerator can be factored
into two sets of parentheses, where the first term is π₯. The second terms in each of the
parentheses must have a product of 36 and a sum of 12. π₯ squared plus 12π₯ plus 36 is
equal to π₯ plus six multiplied by π₯ plus six or π₯ plus six all squared. As we have already factored the
denominator, π of π₯ is equal to π₯ plus six multiplied by π₯ plus six over π₯
minus six multiplied by π₯ plus six. The denominator has zeros at π₯
equals positive and negative six. Therefore, the function is
undefined at these values. This means that the domain of
π of π₯ is the set of all real values minus the set containing negative six and
six. As negative six is not
contained in the domain, we can cancel the factor of π₯ plus six. And the function π of π₯
simplifies to π₯ plus six over π₯ minus six as required. This confirms that the value of
π is 36.

So far in this video, we have only
dealt with quadratic polynomials, but we will now look at one involving a cubic
polynomial.

Simplify the function π of π₯
is equal to π₯ squared minus 81 over π₯ cubed plus 729 and find its domain.

In this question, we will begin
by finding the domain of the function and then simplify it by factoring the
numerator and denominator and canceling any shared factors. We recall that the domain of a
rational function is all real values of π₯ except where the denominator is equal
to zero. We therefore need to set the
cubic polynomial π₯ cubed plus 729 equal to zero. In order to solve this
equation, we will begin by factoring the left-hand side. 729 is a cube number. It is equal to nine cubed. This means that π₯ cubed plus
729 is written in the form π₯ cubed plus π cubed. The sum of cubes formula states
that this is equal to π₯ plus π multiplied by π₯ squared minus ππ₯ plus π
squared. π₯ cubed plus 729 is therefore
equal to π₯ plus nine multiplied by π₯ squared minus nine π₯ plus 81.

At this stage, we have the
product of a linear and quadratic term equal to zero. Setting the linear factor equal
to zero gives us π₯ is equal to negative nine. The quadratic equation π₯
squared minus nine π₯ plus 81 equals zero has no real solutions. This is because the
discriminant π squared minus four ππ is less than zero. We therefore have only one real
solution to the cubic equation π₯ cubed plus 729 equals zero. It is π₯ is equal to negative
nine. As this is the only value of π₯
that makes the function undefined, we can conclude that the domain of π of π₯
is the set of all real values minus the set containing negative nine.

Letβs now move on to
simplifying the function. The numerator is written in the
form π₯ squared minus π squared and can therefore be factored using the
difference of two squares. This is equal to π₯ plus π
multiplied by π₯ minus π. Since the square root of 81 is
nine, π of π₯ is equal to π₯ plus nine multiplied by π₯ minus nine over π₯ plus
nine multiplied by π₯ squared minus nine π₯ plus 81. Since π₯ cannot equal negative
nine, we can cancel the shared factor of π₯ plus nine from the numerator and
denominator. π of π₯ simplifies to π₯ minus
nine over π₯ squared minus nine π₯ plus 81. And we now have answers to the
two parts of the question.

Before moving on to one final
example, we will consider what it means for two rational functions to be equal and
equivalent. If π sub one of π₯ and π sub two
of π₯ are rational functions, we say that π sub one is equal to π sub two if they
have the same domain and are equal on this entire domain. Note that this is the same as
saying the zeros of the denominators of both functions are equal and π sub one
equals π sub two on this domain. The difference between this and the
equivalence of rational functions is as follows. If π sub one of π₯ and π sub two
of π₯ are rational functions, we say that π sub one is equivalent to π sub two if
they are equal on their shared domain. We will now look at an example that
considers this.

Given the functions π sub one
of π₯ is equal to π₯ over π₯ squared minus 10π₯ and π sub two of π₯ is equal to
one over π₯ minus 10, what is the set of values on which π sub one is equal to
π sub two?

In this question, we want to
determine the set of values on which the two given functions are equal. We note that if the functions
are equal on their entire shared domain, then they are equivalent. And since both functions are
rational, we can simplify the functions by finding their domains and canceling
shared factors. Since the numerator and
denominator of π sub two of π₯ have no common factors, this function cannot be
simplified. We will therefore start with
the first function.

We recall that the domain of a
rational function is the set of all real numbers excluding those where the
denominator is equal to zero. Setting π₯ squared minus 10π₯
equal to zero, we can factor out a common factor of π₯ such that π₯ multiplied
by π₯ minus 10 equals zero. This gives us two solutions: π₯
equals zero and π₯ equals 10. The denominator is equal to
zero and the function is undefined when π₯ equals zero and π₯ equals 10. And we can therefore conclude
that the domain of π sub one of π₯ is the set of all real numbers minus the set
containing zero and 10.

Simplifying the denominator of
π sub one of π₯, we can then cancel a shared factor of π₯. This means that π sub one of
π₯ is equal to one over π₯ minus 10 when π₯ is any real number apart from zero
or 10. This is the same expression as
π sub two of π₯. And we can therefore conclude
that the two functions are equal for the set of all real values apart from the
set containing zero and 10. It is worth noting that as the
functions are equal on this set, they are equivalent.

We will now summarize the key
points from this video. We saw in this video that the
domain of a rational function is all real values except those that make the
denominator equal to zero. To simplify a rational function π
of π₯, which is equal to π of π₯ over π of π₯, we find the domain of π of π₯ by
finding the roots of π of π₯. We then fully factor both π of π₯
and π of π₯, cancel the shared factors in the numerator and denominator, and equate
π of π₯ to the simplified expression over the domain of π of π₯. We also saw that if two rational
functions have the same simplified expressions, then theyβre equal across the
intersections of their domains. Finally, if the simplified forms of
two rational functions are equal, then the two functions are equivalent.