Lesson Video: Simplifying Rational Functions | Nagwa Lesson Video: Simplifying Rational Functions | Nagwa

# Lesson Video: Simplifying Rational Functions

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

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### 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, letβs 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 doesnβt 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₯, weβll 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 ππ₯ 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. Letβs 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 ππ₯ squared plus ππ₯ plus π has no real solutions when its discriminant, π squared minus four ππ, 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. Letβs 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. Letβs 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.

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