Solve each problem. This rational function has two holes and o...

Question

Solve each problem. This rational function has two holes and one vertical asymptote. ƒ(x)=(x^3+7x^2-25x-175)/(x^3+3x^2-25x-75)

What are the x-values of the holes?

Accepted Answer

Hello. In this video, we are going to be determining the X values that create holes in the graph of the rational function. So the function that is given to us is F of X is equal to X cubed plus X squared minus 169 X minus 169 is divided by X cubed plus nine X squared minus 169 X minus 1521. In order to start this problem, let's first simplify the given rational function. We are going to start by simplifying the numerator of this function. Now, this numerator is given to us again as X cubed plus X squared minus 169 X minus 169. Since this is a polynomial of four terms, we can simplify this by using factoring by grouping. So we are going to group up the first two and last two terms. And we are going to factor out common factors from each group. If we take a look at the first group, we have X cubed plus X squared. The smallest term we have available to us is X squared. So we are going to factor out an X squared from the first group. In the second group, we have negative 169 minus 169. The common factor of this group is going to be negative 169. If we factor out these terms, we can rewrite the numerator to be X squared multiplied by X plus one minus 169 multiplied by X plus one. Notice that both of these terms have the common group X plus one. What we can do is we can combine the coefficients of these group and we can simplify the numerator to be X squared minus 169 multiplied by X plus one. We can actually factor further notice that X squared minus 169 can be factored by the difference of squares. 169 has roots 13 because 13 is a perfect square of 169. So by using the difference of squares, we can simplify the numerator to be X plus 13 multiplied by X minus 13, multiplied by X plus one. This is the simplest form of the numerator. So now let's go ahead and simplify the denominator. The denominator is given to us as X cubed plus nine X squared minus 169 X minus 1521. And just like the numerator, we are going to simplify this by using factoring by grouping. So we'll group up the first two terms and last two terms together and determine the common factors. So in the first group, the common factor that we can factor out is going to be X squared. And in the second group, we can factor out negative 169 this will allow us to rewrite the denominator as X squared multiplied by X plus nine minus 169 multiplied by X plus nine. We can then combine the coefficients and leave us with X squared minus 169 multiplied by X plus nine. Just like the numerator, we can factor X squared minus 69 by using the difference of squares. And that will allow us to simplify the denominator to be X plus 13 multiplied by X minus 13 multiplied by X plus nine. Now that we have the numerator and denominator simplified, let's replace these into the original function that will give us F of X is equal to X plus 13, multiplied by X minus 13, multiplied by X plus one, divided by X plus 13, multiplied by X minus 13 multiplied by X plus nine. Now, in this form of the function, if we wanted to find the restrictions of the domain that cause the denominator to be zero, we can take each of the factors of X in the denominator, set it equal to zero and solve So what this means is that we get X plus 13 equal to zero, X minus 13 equal to zero and X plus nine equal to zero. And if we solve all three of these equations, we get X is equal to negative 13, X is equal to positive 13 and X is equal to negative nine. These three numbers cannot be included in the domain because they will cause division by zero errors. However, notice that we have factors of X plus 13 and X minus 13 in both the numerator and denominator. What this means is that we can eliminate these factors and remove these restrictions on the denominator. But if we do this, this is going to cause holes in the graph of the function. So what this means is that the removable discontinuities of this function are going to be the values of positive and negative 13 and these are going to be the values of X that cause holes in the function itself. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding on how to remove these discontinuities from the denominator. And I'll go ahead and see you all in the next video.

Solve each problem. This rational function has two holes and one vertical asymptote. ƒ(x)=(x^3+7x^2-25x-175)/(x^3+3x^2-25x-75)

What are the x-values of the holes?

Key Concepts

Rational Functions

Holes in Rational Functions

Vertical Asymptotes

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