### Video Transcript

Simplify the function π of π₯
equals π₯ plus three over π₯ squared minus 49 minus three π₯ cubed plus 21π₯ squared
over three π₯ to the fourth power minus six π₯ cubed minus 105π₯ squared, and
determine its domain.

Letβs begin by inspecting our
function π of π₯ in a little more detail. We can see itβs the difference
between two individual functions. Each of these functions is said to
be a rational function. In other words, itβs a quotient of
a pair of polynomials. Both the numerator and denominator
of each fraction are a polynomial. So in order to simplify this
function, we need to find a way to create common denominators but also cancel any
common factors in the numerators and denominators of our functions.

Before we can do that though, we do
need to determine the domain of π of π₯. So how do we find the domain of a
combination of functions? Well, the domain of a combination
of functions, that is, the sum, difference, or product, is simply the intersection
of the domains of the respective functions. The same holds for when we find the
quotient of a pair of functions. But in this case, a second criteria
must be satisfied. And that is the denominator, the
function that weβre dividing by, cannot be equal to zero.

So we need to find the domain of
each of our rational functions. So letβs define them separately to
begin with. Letβs define our first function to
be π of π₯. And the second function, three π₯
cubed plus 21π₯ squared over three π₯ to the fourth power minus six π₯ cubed minus
105π₯ squared, letβs call that π of π₯. And remember, we said both π of π₯
and π of π₯ are rational functions. And the domain of a rational
function is the set of real numbers, but we exclude any values of π₯ that make the
denominator zero.

Now the reason this is the case is
because a rational function is the quotient of a pair of polynomials. And we said that when we find the
domain of a combination of functions made up of a quotient, the domain is the
intersection of their domains excluding any values that make the function weβre
dividing by zero. And the domain of a polynomial is
the set of real numbers.

So we know the domain of π of π₯
and π of π₯ will be the set of real numbers, but we need to exclude values that
make the denominator in each case equal to zero.

To find these, perhaps somewhat
counterintuitively, weβll actually set the denominators equal to zero and solve for
π₯. Letβs start with the denominator of
π of π₯. Weβre going to solve the equation
π₯ squared minus 49 equals zero. Now there are two ways we can do
this. We can factor by using the
difference of squares, or we can add 49 to both sides. Letβs look at this latter
method.

Adding 49 to both sides gives us π₯
squared equals 49. Then our next job is to find the
square root of both sides. But remember, we have to find both
the positive and negative square root of 49. So π₯ is equal to the positive and
negative square root of 49, or π₯ is equal to positive or negative seven. These are the values of π₯ which we
want to exclude from the domain of π of π₯. So the domain of π of π₯ is the
set of real numbers minus the set containing negative seven and seven.

Letβs repeat this with our second
function π of π₯. Once again, weβll set the
denominator equal to zero. We will have to factor this
expression in order to solve. And to begin with, we might notice
that there is a common factor of π₯ squared throughout each term. In fact, the coefficient of each
term is also a multiple of three. So we can write three π₯ squared on
the outside of our parentheses. Then if we divide each term by
three π₯ squared, the inside of our parentheses becomes π₯ squared minus two π₯
minus 35.

Then we notice that we have a
quadratic expression. And if we can find a pair of terms
that have a product of negative 35 and a sum of negative two, then we can factor
further. Those are negative seven and
five. Negative seven times five is
negative 35, and negative seven plus five is negative two. So we have three π₯ squared times
π₯ minus seven times π₯ plus five equals zero.

And of course for the product of
any three expressions to be equal to zero, we know that at least one of those
expressions must itself be zero. So weβre going to solve three
equations. Weβre going to solve the equation
three π₯ squared equals zero, π₯ minus seven equals zero, and π₯ minus five equals
zero.

For our first equation, if we
divide by three and then take the square root, we get π₯ equals zero. With our second, we add seven to
both sides. So π₯ equals seven. And with our last equation, we take
away five from both sides. So π₯ is negative five. And once again, these are the
values that weβre going to exclude from the domain of our function. So π of π₯ has a domain of the set
of real numbers minus the set containing negative five, zero, and seven.

The domain of our function is the
intersection of these two domains. So the domain of π of π₯ is the
set of real numbers minus all of those values of π₯ which we excluded from both
domains. Thatβs the set containing negative
seven, negative five, zero, and seven.

We still, however, need to simplify
our function. So weβre going to rewrite π of π₯,
but weβll write the denominator of π of π₯ in its factored form as we calculated
earlier. Letβs clear some space for the next
step.

So π of π₯ can be rewritten at
least at first as shown. In order to factor though, we do
need to find some more common factors. So letβs see if thereβs anything
else we can factor. The denominator of our first
fraction can be written as π₯ plus seven times π₯ minus seven. And the numerator of our second can
be written as three π₯ squared times π₯ plus seven.

Now, since weβve excluded zero from
the domain of our function, we know that three π₯ squared divided by three π₯
squared will never be zero divided by zero, which is undefined. And so we can simplify by dividing
this second fraction by three π₯ squared on both the numerator and denominator.

Once weβve done this, we see we can
create a common denominator between our two fractions. That common denominator is going to
be π₯ plus seven times π₯ minus seven times π₯ plus five. And so to achieve this with our
first fraction, weβll need to multiply by π₯ plus five. And of course, weβll do this to the
numerator. Similarly, in our second fraction,
weβll need to multiply by π₯ plus seven. So weβll do that to our
numerator.

And so weβve now created a pair of
fractions with equivalent denominators. π of π₯ is then π₯ plus three
times π₯ plus five over π₯ minus seven times π₯ plus seven times π₯ plus five minus
π₯ plus seven times itself over π₯ minus seven times π₯ plus seven times π₯ plus
five. And then we can simply subtract the
numerators now that the denominator is the same.

Then to see whatβs happening on the
numerator, letβs distribute our parentheses, beginning with π₯ plus three times π₯
plus five. Distributing these with whatever
method we feel comfortable using, we get π₯ squared plus eight π₯ plus 15. And then π₯ plus seven times itself
is π₯ squared plus 14π₯ plus 49.

Weβre going to subtract this second
expression from π₯ squared plus eight π₯ plus 15. When we do, we notice that π₯
squared minus π₯ squared is zero. Eight π₯ minus 14π₯ is negative six
π₯. And 15 minus 49 is negative 34. We might then choose to take out a
common factor of negative two to give us our numerator as negative two times three
π₯ plus 17.

And so we fully simplified our
function π of π₯. And we found its domain. π of π₯ is negative two times
three π₯ plus 17 over π₯ minus seven times π₯ plus seven times π₯ plus five. And its domain is the set of real
numbers minus the set containing negative seven, negative five, zero, and seven.