In Exercises 7–14, simplify each rational expression. Find all nu...

Question

In Exercises 7–14, simplify each rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression. (x^2−14x+49)/(x^2−49)

Accepted Answer

Welcome back. I'm so glad you're here. We've got this rational expression. I'll rewrite it. We've got X squared minus eight X plus 15 in the numerator. All over X squared minus five X plus six in the denominator and we're going to do a couple of things. We're going to write all the numbers that make this expression undefined. And we are going to simplify, we're actually going to go back and forth, we're going to start doing some simplification and then we're going to pause, find out what numbers make it undefined and then go back to our final simplification. So the first step of simplification is to factor everything as much as possible. Let's look at the numerator. If we want to factor that we say what times what gives us X squared? Well that's going to be X times X. And then what times what multiplies to be positive 15 but adds up to be a negative eight and while that's going to be a negative times a negative so that it multiplies to be positive but adds up to be negative and that's going to be a three and a five because negative three plus negative five equals negative eight. A negative three times negative five equals positive 15. Great, that's the numerator factor. Now it's factor the denominator. What times what gives us X squared. Well that's X times X. And now what times what gives us a positive six but adds up to be a negative five again. This is going to be a negative times a negative so that it equals a positive but adds up to be negative and it's going to be a two and a three because a negative two plus a negative three gives us a negative five and negative two times negative three gives us a positive six. So that's as factored as we can get that. And here is where we pause and we have to ask ourselves what would make this expression undefined? Why do we have to do it right here? Why can't we keep going with the next step of simplification? Which is to cancel out factors that occur in the numerator and the denominator because we might inadvertently cancel out something that would make it undefined. So we have to pause, figure out what makes it undefined and what is undefined undefined is any time you have a zero in the denominator you can't do that, you can't divide by zero. So in order to find out why, how we would have a zero in the denominator, it's like we have to sneak and we have to set these two factors equal to zero. So X minus two equals zero, add two to both sides. So we get X equals two and x minus three equals zero will add three double size. That gives us X equals three. And what did that tell us that tells us what X could equal. That would make it zero in the denominator. Therefore X cannot equal those numbers X cannot be two or three great sneaky work. Now we know what makes it undefined and we can go back to our beautiful simplification here and what factors occur in both the numerator and the denominator. This x minus three. Get out of here, you are canceled, you become a one and a one. Now we've got one times X minus five, that's x minus five in the numerator. And we've got x minus two times one which is x minus two in the denominator. That's as simplified as it's going to get. So we look at our answer choices and that matches answer choice. D. Well done, we'll catch you on the next one.

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