Multiplying & dividing rational expressions
Multiply and express as a simplified rational. State the domain. Let's multiply it, and then before we simplify it, let's look at the domain. This is equal to, if we just multiplied the numerators, a squared minus 4 times a plus 1, all of that over-- multiply the denominators-- a squared minus 1 times a plus 2. Now, the a squared minus 4 and the a squared minus 1 might look familiar to us. These are the difference in squares, a special type of binomial that you could immediately, or hopefully maybe immediately recognize. It takes the form a squared minus b squared, difference of squares, and it's always going to be equal to a plus b times a minus b. We can factor this a squared minus 4 and we can also factor this a squared minus 1, and that'll help us actually simplify the expression or simplify the rational. This top part, we can factor the a squared minus 4 as a plus 2-- 2 squared is 4-- times a minus 2, and then that times a plus 1. Then in the denominator, we can factor a squared minus 1-- let me do that in a different color. a squared minus 1 we can factor as a plus 1 times a minus 1. If you ever want to say, hey, why does this work? Just multiply it out and you'll see that when you multiply these two things, you get that thing right there. Then in the denominator, we also have an a plus 2. We've multiplied it, we've factored out the numerator, we factored out the denominator. Let's rearrange it a little bit. So this numerator, let's put the a plus 2's first in both the numerator and the denominator. So this, we could get a plus 2 in the numerator, and then in the denominator, we also have an a plus 2. In the numerator, we took care of our a plus 2's. That's the only one that's common, so in the numerator, we also have an a minus 2. Actually, we have an a plus 1-- let's write that there, too. We have an a plus 1 in the numerator. We also have an a plus 1 in denominator. In the numerator, we have an a minus 2, and in the denominator, we have an a minus 1. So all I did is I rearranged the numerator and the denominator, so if there was something that was of a similar-- if the same expression was in both, I just wrote them on top of each other, essentially. Now, before we simplify, this is a good time to think about the domain or think about the a values that aren't in the domain, the a values that would invalidate or make this expression undefined. Like we've seen before, the a values that would do that are the ones that would make the denominator equal 0. So the a values that would make that equal to 0 is a is equal to negative 2. You could solve for i. You could say a plus 2 is equal to 0, or a is equal to negative 2. a plus 1 is equal to 0. Subtract 1 from both sides. a is equal to negative 1. Or a minus 1 is equal to 0. Add one to both sides, and you get a is equal to 1. For this expression right here, you have to add the constraint that a cannot equal negative 2, negative 1, or 1, that a can be any real number except for these. We're essentially stating our domain. We're stating the domain is all possible a's except for these things right here, so we'd have to add that little caveat right there. Now that we've done that, we can factor it. We have an a plus 2 over an a plus 2. We know that a is not going to be equal to negative 2, so that's always going to be defined. When you divide something by itself, that is going to just be 1. The same thing with the a plus 1 over the a plus 1. That's going to be 1. All you're going to be left with is an a minus 2 over a minus 1. So the simplified rational is a minus 2 over a minus 1 with the constraint that a cannot equal negative 2, negative 1, or 1. You're probably saying, Sal, what's wrong with it equaling, for example, negative 1 here? Negative 1 minus 1, it's only going to be a negative 2 here. It's going to be defined. But in order for this expression to really be the same as this expression up here, it has to have the same constraints. It has to have the same domain. It cannot be defined at negative 1 if this guy also is not defined at negative 1. And so these constraints essentially ensure that we're dealing with the same expression, not one that's just close.