### Video Transcript

In this video, we will learn how to
simplify rational functions and how to find their domains. We will simplify rational functions
by finding and eliminating common factors in the numerators and denominators. We will do this by factoring. We will also work out the domain of
a rational function by considering values of π₯ for which the denominator is equal
to zero. We will begin by looking at some
key definitions.

A rational function is any function
which can be defined by a rational fraction. In this video, we will look at
algebraic fractions where both the numerator and denominator are polynomials. The domain of a function π of π₯
is the set of π₯-values we can input into the function. There will be no real solutions to
a function when the denominator is equal to zero. We can therefore set the
denominator equal to zero to calculate the real values that are not in the
domain.

For example, letβs consider the
function four π₯ plus seven over π₯ minus three. This function is already written in
its simplest form. As there are no obvious
restrictions, we might think that the domain is all real values. However, as the denominator cannot
be equal to zero, we need to set this equal to zero to work out the values that are
not in the domain. Adding three to both sides of this
equation gives us π₯ is equal to three. This means that when π₯ equals
three, the denominator will be zero. This means that the domain of our
function is all real values minus the set containing the number three.

We will now look at some questions
where we need to simplify rational functions and determine their domain.

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

The first part of the question asks
us to simplify the function. We will do this by factoring or
factorizing the numerator and denominator. The numerator has a common factor
of π₯, so we can factor this out. As π₯ squared divided by π₯ is π₯
and two π₯ divided by π₯ is two, the numerator becomes π₯ multiplied by π₯ plus
two. We could check this by
redistributing the parentheses. The denominator of our fraction is
written in the form π₯ squared minus π squared. This is the difference of two
squares and is equal to π₯ plus π multiplied by π₯ minus π. π₯ squared minus four is equal to
π₯ plus two multiplied by π₯ minus two.

Once again, we could check this by
distributing the parentheses using the FOIL method. We notice that we have π₯ plus two
on the numerator and the denominator. This means that we can cancel
these. This gives us a simplified version
of π of π₯ equal to π₯ divided by π₯ minus two.

We are also asked to find the
domain of this function, π of π₯. The domain is the π₯-values that we
can input into π of π₯. It doesnβt appear to be any obvious
restriction here; however, we recall that the denominator of any fraction cannot be
equal to zero. We therefore need to solve the
denominator π₯ squared minus four equals zero to identify the values that are not in
the domain of π of π₯.

As we have already seen, π₯ squared
minus four can be rewritten as π₯ plus two multiplied by π₯ minus two. If the product of these parentheses
is equal to zero, then either π₯ plus two equals zero or π₯ minus two equals
zero. This gives us two solutions: π₯
equals negative two and π₯ equals two. The domain of π of π₯ is therefore
equal to all real values apart from the set negative two and two.

Our next question involves finding
the value of an unknown in a rational 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?

There are lots of ways of
approaching this problem. One way would be to set both of the
expressions or functions equal to each other. This gives us π₯ squared plus 12π₯
plus 36 over π₯ squared minus π is equal to π₯ plus six over π₯ minus six. The numerator on the left-hand side
is a quadratic that can be factorized into two parentheses or brackets. As the leading term has a
coefficient of one, the first term in each of our brackets will be π₯. We need to find a pair of numbers
that have a product of 36 and a sum of 12.

There are five factor pairs of 36:
one and 36, two and 18, three and 12, four and nine, and six and six. The only pair that sum to 12 is six
and six. The factorized form of π₯ squared
plus 12π₯ plus 36 is therefore equal to π₯ plus six multiplied by π₯ plus six. Our next step is to divide both
sides of the equation by π₯ plus six. This gives us π₯ plus six over π₯
squared minus π is equal to one over π₯ minus six. We can then cross multiply. We can multiply both sides of the
equation by π₯ squared minus π and π₯ minus six. This gives us π₯ plus six
multiplied by π₯ minus six on the left-hand side and π₯ squared minus π on the
right-hand side.

We could expand the left-hand side
using the FOIL method, multiplying the first, outside, inside, and last terms. This would give us π₯ squared minus
six π₯ plus six π₯ minus 36. The negative six π₯ and positive
six π₯ here cancel. Alternatively, we might have
noticed that π₯ plus six multiplied by π₯ minus six was the difference of two
squares and was therefore equal to π₯ squared minus 36. We know that this is equal to π₯
squared minus π. Our value of π is therefore equal
to 36.

Our next rational function has a
denominator that is a cubic.

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

In order to simplify the function
π of π₯, weβll need to factorize the numerator and denominator. In order to do this, we need to
recognize that on the numerator, we have the difference of two squares. This tells us that any quadratic of
the form π₯ squared minus π squared is equal to π₯ plus π multiplied by π₯ minus
π. We know that nine squared is equal
to 81. This means that π₯ squared minus 81
is equal to π₯ plus nine multiplied by π₯ minus nine. The denominator π₯ cubed plus 729
is written in the form π₯ cubed plus π cubed.

Once again, π is equal to nine as
nine cubed is 729. π₯ cubed plus π cubed is equal to
π₯ plus π multiplied by π₯ squared minus ππ₯ plus π squared. This means that π₯ cubed plus 729
is equal to π₯ plus nine multiplied by π₯ squared minus nine π₯ plus 81. We can divide the numerator and
denominator by π₯ plus nine, so these terms cancel. The simplified version of π of π₯
is therefore equal to π₯ minus nine over π₯ squared minus nine π₯ plus 81.

The second part of our question
asks us to find the domain. The domain of a function is the set
of π₯-values we can substitute or input into π of π₯. As the denominator of a fraction
cannot be equal to zero, the domain will be all real values apart from those which
make the denominator zero. To calculate these values, we set
π₯ cubed plus 729 equal to zero. We already know this is equal to π₯
plus nine multiplied by π₯ squared minus nine π₯ plus 81. Letβs consider the linear term π₯
plus nine first. When π₯ plus nine is equal to zero,
π₯ is equal to negative nine. Therefore, negative nine cannot be
in the domain of π of π₯. The quadratic π₯ squared minus nine
π₯ plus 81 cannot be factored or factorized.

We can actually go one stage
further here as it actually has no real solutions. We know that any quadratic of the
form ππ₯ squared plus ππ₯ plus π has no real solutions when its discriminant, π
squared minus four ππ, is less than zero. In this case, we have π is equal
to one, the coefficient of π₯ squared, π is equal to negative nine, the coefficient
of π₯, and π is 81. We need to calculate negative nine
squared minus four multiplied by one multiplied by 81. This is equal to negative 243. As this is less than naught, the
quadratic π₯ squared minus nine π₯ plus 81 equals zero has no solutions.

This means that the only value such
that π₯ cubed plus 729 equals zero is π₯ is equal to negative nine. The domain of π of π₯ is therefore
equal to all real values apart from negative nine. We can write this in set notation
as shown.

Our final question involves
simplifying a more complicated rational function.

Simplify the function π of π₯ is
equal to five π₯ squared minus 15π₯ over π₯ to the fourth power plus two π₯ cubed
minus 15π₯ squared minus 36 minus π₯ squared over π₯ squared minus π₯ minus 30, then
find the solution set of the equation π of π₯ is equal to zero.

We can simplify the function π of
π₯ by factorizing each part of the function. Letβs begin by considering five π₯
squared minus 15π₯. This has a highest common factor of
five π₯. Factorizing this out gives us five
π₯ multiplied by π₯ minus three. The denominator of our first term
looks more complicated as it is a quartic function. It has a degree of four, as the
highest exponent is four. There is, however, a common factor
of π₯ squared. π₯ to the fourth power plus two π₯
cubed minus 15π₯ squared is equal to π₯ squared multiplied by π₯ squared plus two π₯
minus 15.

The quadratic part of this can be
factorized further. π₯ squared plus two π₯ minus 15 is
equal to π₯ plus five multiplied by π₯ minus three. This means that the first term has
a numerator of five π₯ multiplied by π₯ minus three and the denominator of π₯
squared multiplied by π₯ plus five multiplied by π₯ minus three. Letβs now consider the numerator of
the second term. 36 minus π₯ squared is in the form
of the difference of two squares, π squared minus π₯ squared. This means that it can be
factorized in the form π minus π₯ multiplied by π plus π₯. 36 minus π₯ squared is equal to six
minus π₯ multiplied by six plus π₯. Finally, the denominator of the
second term simplifies to π₯ minus six multiplied by π₯ plus five.

We can now rewrite the function π
of π₯ in its simplified form. The function π of π₯ is equal to
five π₯ multiplied by π₯ minus three over π₯ squared multiplied by π₯ plus five
multiplied by π₯ minus three minus six minus π₯ multiplied by six plus π₯ over π₯
minus six multiplied by π₯ plus five. In our first term, we can cancel an
π₯ and an π₯ minus three on the numerator and denominator. This gives us five over π₯
multiplied by π₯ plus five. The negative of six minus π₯ is the
same as negative six plus π₯. This, in turn, can be written as π₯
minus six. We can therefore cancel π₯ minus
six on the numerator and denominator of the second term. This leaves us with six plus π₯ or
π₯ plus six over π₯ plus five.

We can now add the two terms of our
function by finding a common denominator. We do this by multiplying the
numerator and denominator of the second fraction by π₯. This gives us π₯ multiplied by π₯
plus six over π₯ multiplied by π₯ plus five. Adding the numerators gives us five
plus π₯ multiplied by π₯ plus six over π₯ multiplied by π₯ plus five. Our next step is to distribute the
parentheses on the numerator. Rewriting this gives us π₯ squared
plus six π₯ plus five. This, in turn, can be factored into
π₯ plus five multiplied by π₯ plus one. Canceling π₯ plus five gives us the
simplified version of π of π₯ of π₯ plus one over π₯.

We also want the solution set where
π of π₯ is equal to zero. This means that π₯ plus one over π₯
equals zero. Multiplying both sides by π₯, we
have π₯ plus one is equal to zero. π₯ is therefore equal to negative
one. The solution set of the equation π
of π₯ equals zero is the value negative one.

We will know summarize the key
points from this video. A function in the form of an
algebraic fraction can be simplified by factoring the numerator and denominator and
then canceling like terms. A function is undefined when the
denominator is equal to zero. The values at which the denominator
is equal to zero will not be contained in the domain. We saw in this video that we can
write the domain using set notation as shown.