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CCSS.Math: ,

multiply and express as a simplified rational state the domain so let's multiply it and then before we simplify it let's look at the domain so this is equal to if we just multiply the numerators a squared minus 4 times a plus 1 a plus 1 all of that over multiply the denominators a squared minus 1 times a plus 2 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 can immediately or hopefully maybe immediately recognize take 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 so 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 the simplify the rational so this top part we can break up the we can factor the a squared minus 4 as a plus 2 right 2 squared is 4 times a minus 2 and then that times a plus 1 times a plus 1 and then in the denominator we can factor a squared minus 1 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 and 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 the denominator we also have an A plus 2 we also have an A plus 2 so we factored the new we've multiplied it we 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 so we took care of our a plus twos we have let's see we have well that's the only one that's common so in the numerator we an a-minus to actually have an a-plus one let's write that there do a plus one we have an A plus 1 in the numerator we also have an A plus 1 in the denominator a plus 1 in the denominator and the numerator we have an a minus 2 a minus 2 and the denominator we have an a minus 1 a minus 1 so all I did is I rearranged the numerator in the denominator so if there was something that it was of a similar it was 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 so like we've seen before the a values that would do that are the ones that would make the denominator equal zero so the a values that would make that equal to zero is a is equal to negative two I mean you could solve for you could say a plus two is equal to zero or a is equal to negative two a plus one is equal to zero subtract one from both sides a is equal to negative 1 or a minus one is equal to zero let's add 1 to both sides and you get a is equal to one so this expression for this expression right here you have to add the constraint that a cannot equal a cannot equal negative two negative one or one that a can be any real number except for these this is this is we're essentially stating our domain we're saying it can be the domain is all possible A's except for these things right here so we'd have to add that little cut that little caveat right there now now that we've done that we can factor it so we have an A plus two over na plus two we know that a is not going to be equal to negative two so that's always going to be defined and so when you divide something by itself that is going to just be one same thing where the eight plus one over the a plus one that's going to be one and all you're going to be left with is an a minus two over a minus one so the simplified rational is a minus two over a minus one with the constraint with the constraint that a cannot equal negative two negative one or one now you're probably saying hey Sal what's wrong with it equally so for example negative one here negative one minus one it's only going to be a negative two here it's going to be defined but for this expression in order for that 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 one if this guy also is not defined at negative one and so this one these constraints essentially ensure that we're dealing with the same expression not one that's just close