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CCSS.Math: ,

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.