Video Transcript
Simplify the function 𝑛 of 𝑥 equals five 𝑥 squared minus 15𝑥 over 𝑥 to the power
of four plus two 𝑥 cubed minus 15𝑥 squared minus 36 minus 𝑥 squared over 𝑥
squared minus 𝑥 minus 30, then find the solution set of the equation 𝑛 of 𝑥
equals zero.
For this question, we have been asked to simplify a function which involves a
subtraction of two rational expressions. In this case, the process will involve a lot of factoring and a lot of canceling. Our first step will be to factor the polynomial expressions which form the numerators
and denominators of the two algebraic fractions.
Let’s start by considering the numerator of our first fraction, five 𝑥 squared minus
15𝑥. The highest common factor of the terms in this expression is five 𝑥. Factoring this out gives us the expression five 𝑥 multiplied by 𝑥 minus three. It is worth nothing that this expression is now made up entirely of linear
factors. These linear factors cannot be factorized further and so can be treated a bit like
the building blocks for a polynomial expression.
For our next few steps, we’ll also aim to express the other polynomials in terms of
linear factors, starting with 𝑥 to the power of four plus two 𝑥 cubed minus 15𝑥
squared. At first, this might seem a bit complicated, since the highest exponent of 𝑥 is
four, making it a quartic polynomial. We might notice, however, that each of the terms has a common factor of 𝑥
squared. This allows us to factor the expression as 𝑥 squared multiplied by 𝑥 squared plus
two 𝑥 minus 15. The quadratic part of this expression can be factored further. For this video, we won’t go into detail on the various methods on how to do this, but
in this case, we should be able to spot that the correct factorization is 𝑥 plus
five multiplied by 𝑥 minus three.
The denominator of our first term is therefore 𝑥 squared multiplied by 𝑥 plus five
multiplied by 𝑥 minus three. Once again, here we have essentially formed an expression comprised of linear
factors. One small note is that 𝑥 squared is not technically a linear factor, but it can be
thought of as 𝑥 times 𝑥. In this sense, it is actually two linear factors, or a repeated linear factor of
𝑥.
Let’s move on to the numerator of our second rational expression, which is 36 minus
𝑥 squared. This is in the form of a difference of two squares, which is 𝑎 squared minus 𝑥
squared. We know that it can therefore be factored in the form 𝑎 minus 𝑥 multiplied by 𝑎
plus 𝑥. In our case, we should recognize that six squared is 36. And so, our expression can be factorized as six minus 𝑥 multiplied by six plus
𝑥.
Okay, finally, we’ll consider the denominator of our second rational expression. Once again, we are faced with a quadratic. And with a bit of inspection, we can see that 𝑥 squared minus 𝑥 minus 30 is equal
to 𝑥 minus six multiplied by 𝑥 plus five.
Now that our numerators and denominators have been expressed in terms of linear
factors, we can now rewrite the function 𝑛 of 𝑥 using our factored
expressions. You might be able to see the substitutions more clearly if we label the numerator and
denominator of the first rational expression as one and two, and the numerator and
denominator of the second rational expression as three and four. Before we move on, let’s clear a bit of space. We’ll also move our new expression for 𝑛 of 𝑥 to the top of the screen.
In order to proceed with the simplification, our next step is to look for common
factors on the top and bottom of our rational expressions that can be canceled
out. Before we do this, however, there is an important intermediate step. Canceling common factors on the top and bottom of a rational expression removes
information about the domain of the underline function. So, we’ll need to investigate the domain before performing the cancelations. We won’t go into too much detail about the domain of a function here, but it’s enough
to remember that a function is undefined at values of 𝑥 where the denominator of
any rational expression is equal to zero. Of course, this is because we cannot divide by zero.
To find these problematic values of 𝑥, we can look at both of our factorized
denominators. When any of the linear factors in our denominators is equal to zero, this in turn
will mean one of the denominators themselves is equal to zero. This situation occurs when either 𝑥 squared equals zero, 𝑥 plus five equals zero,
𝑥 minus three equals zero, or 𝑥 minus six equals zero. If we solve these simple equations, we find that our function 𝑛 of 𝑥 is undefined
when 𝑥 equals negative five, zero, three, and six. Let’s put this information to one side for later and return to our simplification
steps.
As mentioned earlier, we are now going to look for common factors to cancel on the
top and bottom of our rational expressions. In our first rational expression, we can cancel a factor of 𝑥 and a factor of 𝑥
minus three from the top and bottom. For our second rational expression, before we start, we will first reexpress one of
our factors to make the cancelation more obvious. First, note that we are subtracting the second term in the function 𝑛 of 𝑥. This can be thought of as having a factor of negative one shown in front of the
entire term. We also have a factor of six minus 𝑥 in the numerator of this term. Negative one multiplied by six minus 𝑥 can instead be expressed as 𝑥 minus six.
Let’s substitute this back into 𝑛 of 𝑥. We are now adding our second rational expression term, which is 𝑥 minus six
multiplied by six plus 𝑥 over 𝑥 minus six multiplied by 𝑥 plus five. It is now far easier to see that the common factor of 𝑥 minus six can be canceled
from the top and bottom half of the fraction. Great! Our newly simplified expression is therefore five over 𝑥 multiplied by 𝑥 plus five
plus 𝑥 plus six over 𝑥 plus five.
To proceed with the simplification, our aim is now to express the addition of these
two rational expressions as one single rational expression. We do this by creating a common denominator across the two terms. In our case, the simplest way to do this is multiplying the top and bottom of our
second term by 𝑥. This gives us a common denominator of 𝑥 multiplied by 𝑥 plus five. We can now add our two terms to give five plus 𝑥 multiplied by 𝑥 plus six divided
by our new common denominator, 𝑥 multiplied by 𝑥 plus five.
As we said earlier, whenever we have a polynomial expression, it is useful to express
it in terms of linear factors. Currently, our numerator is not in this form. To simplify this numerator, we’ll first distribute the factor of 𝑥 over our
parentheses. And then, we’ll rearrange our expression into the familiar form of a quadratic. Factoring this quadratic, we get 𝑥 plus one multiplied by 𝑥 plus five, which can be
substituted back into our expression for 𝑛 of 𝑥.
Note that we have already considered the domain of our function, so now we can
proceed directly to canceling common factors again. Here, it should be easy to spot that a common factor of 𝑥 plus five can be canceled
from the top and bottom of our fraction. This leaves us with the rational expression of 𝑥 plus one over 𝑥. Since there are no more useful simplifications, we have reached the end of our
process. The function 𝑛 of 𝑥 simplifies to 𝑥 plus one over 𝑥.
Although we’ve reached the end of our simplification, we are not quite finished with
the question yet. Let’s clear some space to consider the final part, which is finding the solution set
of the equation 𝑛 of 𝑥 equals zero. And now that we have a simplified expression for 𝑛 of 𝑥, to find the solution set,
we can simply set this expression equal to zero and solve. In order for the rational expression 𝑥 plus one over 𝑥 to be equal to zero, the
numerator 𝑥 plus one must be equal to zero. We could also think of this as multiplying both sides of our equation by 𝑥. The next step is subtracting one from both sides of the equation. Doing this, we find that 𝑥 is equal to negative one.
For our very final step, let’s quickly check that 𝑥 equals negative one is not one
of the values at which our function 𝑛 of 𝑥 is undefined. If this were true, it would not be part of the solution set. Since our function 𝑛 of 𝑥 is defined when 𝑥 equals negative one, we can say the
solution set is indeed negative one. With this step, we have now found the answer to our question. The simplified expression for 𝑛 of 𝑥 is 𝑥 plus one over 𝑥. And the solution set of the equation 𝑛 of 𝑥 equals zero is negative one.
