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Lesson Video: Simplifying Rational Functions Mathematics • 10th Grade

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

17:52

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 𝑓 of π‘₯, which is equal to 𝑝 of π‘₯ over π‘ž of π‘₯, is a rational function if both 𝑝 of π‘₯ and π‘ž of π‘₯ are polynomials, where π‘ž 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 𝑓 of π‘₯, which is equal to 𝑝 of π‘₯ over π‘ž of π‘₯, is a rational function, then the domain of 𝑓 of π‘₯ is all real numbers except those where π‘ž 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. Let’s consider the function 𝑓 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 𝑓 of π‘₯. Adding six to both sides of our equation, we have π‘₯ equals six. And we can therefore conclude that the domain of 𝑓 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 𝑓 of π‘₯ is equal to 𝑝 of π‘₯ over π‘ž of π‘₯, we follow the following steps. Firstly, we find all the values of π‘₯ where π‘ž of π‘₯ is equal to zero, as the domain of 𝑓 of π‘₯ is all real values except these roots. Our second step is to fully factor both 𝑝 of π‘₯ and π‘ž 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 𝑓 of π‘₯. Finally, we equate 𝑓 of π‘₯ to the simplified expression restricted to the domain of 𝑓 of π‘₯. We will now consider how we can apply this process to an example.

Simplify the function 𝑓 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 𝑓 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‘ž 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 𝑓 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 we’ve 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. 𝑓 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 we’ve already established that π‘₯ equals negative two is not in the domain of 𝑓 of π‘₯. And the function simplifies to π‘₯ over π‘₯ minus two.

We now have the two answers to this question. The function 𝑓 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 𝑛 of π‘₯ is equal to π‘₯ squared plus 12π‘₯ plus 36 over π‘₯ squared minus π‘Ž simplifies to 𝑛 of π‘₯ is equal to π‘₯ plus six over π‘₯ minus six, what is the value of π‘Ž?

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‘Ž 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 𝑦 squared. And this can be factored using the difference of two squares into two linear factors: π‘₯ minus 𝑦, π‘₯ plus 𝑦. Since 𝑦 is equal to six, we have π‘₯ squared minus π‘Ž is equal to π‘₯ minus six multiplied by π‘₯ plus six. And distributing the parentheses or noting that π‘Ž is equal to 𝑦 squared, the value of π‘Ž is equal to 36.

We can verify this by considering the initial expression for 𝑛 of π‘₯. If π‘Ž is equal to 36, 𝑛 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, 𝑛 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 𝑛 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 𝑛 of π‘₯ simplifies to π‘₯ plus six over π‘₯ minus six as required. This confirms that the value of π‘Ž 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 𝑓 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‘Ž cubed. The sum of cubes formula states that this is equal to π‘₯ plus π‘Ž multiplied by π‘₯ squared minus π‘Žπ‘₯ plus π‘Ž 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 𝑏 squared minus four π‘Žπ‘ 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 𝑓 of π‘₯ is the set of all real values minus the set containing negative nine.

Let’s now move on to simplifying the function. The numerator is written in the form π‘₯ squared minus π‘Ž squared and can therefore be factored using the difference of two squares. This is equal to π‘₯ plus π‘Ž multiplied by π‘₯ minus π‘Ž. Since the square root of 81 is nine, 𝑓 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. 𝑓 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 𝑛 sub one of π‘₯ and 𝑛 sub two of π‘₯ are rational functions, we say that 𝑛 sub one is equal to 𝑛 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 𝑛 sub one equals 𝑛 sub two on this domain. The difference between this and the equivalence of rational functions is as follows. If 𝑛 sub one of π‘₯ and 𝑛 sub two of π‘₯ are rational functions, we say that 𝑛 sub one is equivalent to 𝑛 sub two if they are equal on their shared domain. We will now look at an example that considers this.

Given the functions 𝑛 sub one of π‘₯ is equal to π‘₯ over π‘₯ squared minus 10π‘₯ and 𝑛 sub two of π‘₯ is equal to one over π‘₯ minus 10, what is the set of values on which 𝑛 sub one is equal to 𝑛 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 𝑛 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 𝑛 sub one of π‘₯ is the set of all real numbers minus the set containing zero and 10.

Simplifying the denominator of 𝑛 sub one of π‘₯, we can then cancel a shared factor of π‘₯. This means that 𝑛 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 𝑛 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 𝑓 of π‘₯, which is equal to 𝑝 of π‘₯ over π‘ž of π‘₯, we find the domain of 𝑓 of π‘₯ by finding the roots of π‘ž of π‘₯. We then fully factor both 𝑝 of π‘₯ and π‘ž of π‘₯, cancel the shared factors in the numerator and denominator, and equate 𝑓 of π‘₯ to the simplified expression over the domain of 𝑓 of π‘₯. We also saw that if two rational functions have the same simplified expressions, then they’re equal across the intersections of their domains. Finally, if the simplified forms of two rational functions are equal, then the two functions are equivalent.

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