# Question Video: Determining the Domain of the Difference of Two Rational Functions by Simplifying Mathematics

Simplify the function 𝑛(𝑥) = ((𝑥 + 3)/(𝑥² − 49)) − ((3𝑥³ + 21𝑥²)/(3𝑥⁴ − 6𝑥³ − 105𝑥²)), and determine its domain.

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### Video Transcript

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

Let’s begin by inspecting our function 𝑛 of 𝑥 in a little more detail. We can see it’s the difference between two individual functions. Each of these functions is said to be a rational function. In other words, it’s 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 𝑛 of 𝑥. 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 we’re dividing by, cannot be equal to zero.

So we need to find the domain of each of our rational functions. So let’s define them separately to begin with. Let’s define our first function to be 𝑓 of 𝑥. And the second function, three 𝑥 cubed plus 21𝑥 squared over three 𝑥 to the fourth power minus six 𝑥 cubed minus 105𝑥 squared, let’s call that 𝑔 of 𝑥. And remember, we said both 𝑓 of 𝑥 and 𝑔 of 𝑥 are rational functions. And the domain of a rational function is the set of real numbers, but we exclude any values of 𝑥 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 we’re dividing by zero. And the domain of a polynomial is the set of real numbers.

So we know the domain of 𝑓 of 𝑥 and 𝑔 of 𝑥 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, we’ll actually set the denominators equal to zero and solve for 𝑥. Let’s start with the denominator of 𝑓 of 𝑥. We’re going to solve the equation 𝑥 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. Let’s look at this latter method.

Adding 49 to both sides gives us 𝑥 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 𝑥 is equal to the positive and negative square root of 49, or 𝑥 is equal to positive or negative seven. These are the values of 𝑥 which we want to exclude from the domain of 𝑓 of 𝑥. So the domain of 𝑓 of 𝑥 is the set of real numbers minus the set containing negative seven and seven.

Let’s repeat this with our second function 𝑔 of 𝑥. Once again, we’ll 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 𝑥 squared throughout each term. In fact, the coefficient of each term is also a multiple of three. So we can write three 𝑥 squared on the outside of our parentheses. Then if we divide each term by three 𝑥 squared, the inside of our parentheses becomes 𝑥 squared minus two 𝑥 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 𝑥 squared times 𝑥 minus seven times 𝑥 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 we’re going to solve three equations. We’re going to solve the equation three 𝑥 squared equals zero, 𝑥 minus seven equals zero, and 𝑥 minus five equals zero.

For our first equation, if we divide by three and then take the square root, we get 𝑥 equals zero. With our second, we add seven to both sides. So 𝑥 equals seven. And with our last equation, we take away five from both sides. So 𝑥 is negative five. And once again, these are the values that we’re going to exclude from the domain of our function. So 𝑔 of 𝑥 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 𝑛 of 𝑥 is the set of real numbers minus all of those values of 𝑥 which we excluded from both domains. That’s the set containing negative seven, negative five, zero, and seven.

We still, however, need to simplify our function. So we’re going to rewrite 𝑛 of 𝑥, but we’ll write the denominator of 𝑔 of 𝑥 in its factored form as we calculated earlier. Let’s clear some space for the next step.

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

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

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

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

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

We’re going to subtract this second expression from 𝑥 squared plus eight 𝑥 plus 15. When we do, we notice that 𝑥 squared minus 𝑥 squared is zero. Eight 𝑥 minus 14𝑥 is negative six 𝑥. 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 𝑥 plus 17.

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