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CCSS.Math: ,

Find the difference. Express the answer as a
simplified rational expression, and state
the domain. We have two rational
expressions, and we're subtracting one from
the other. Just like when we first learned
to subtract fractions, or add fractions, we have to
find a common denominator. The best way to find a common
denominator, if were just dealing with regular numbers, or
with algebraic expressions, is to factor them out, and
make sure that our common denominator has all of the
factors in it-- that'll ensure that it's divisible by the
two denominators here. This guy right here is
completely factored-- he's just a plus 2. This one over here, let's see if
we can factor it: a squared plus 4a plus 4. Well, you see the pattern that
4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4
is a plus 2 times a plus 2, or a plus 2 squared. We could say it's a plus 2 times
a plus 2-- that's what a squared plus 4a plus 4 is. This is obviously divisible
by itself-- everything is divisible by itself, except, I
guess, for 0, is divisible by itself, and it's also divisible
by a plus 2, so this is the least common multiple of
this expression, and that expression, and it could be
a good common denominator. Let's set that up. This will be the same thing as
being equal to this first term right here, a minus 2 over
a plus 2, but we want the denominator now to be a plus 2
times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this
numerator and denominator by a plus 2, so its denominator is
the same thing as this. Let's multiply both the
numerator and the denominator by a plus 2. We're going to assume that a
is not equal to negative 2, that would have made this
undefined, and it would have also made this undefined. Throughout this whole thing,
we're going to assume that a cannot be equal to negative 2. The domain is all real numbers,
a can be any real number except for negative 2. So, the first term is that--
extend the line a little bit-- and then the second term doesn't
change, because its denominator is already the
common denominator. Minus a minus 3 over-- and we
could write it either as a plus 2 times a plus 2, or
as this thing over here. Let's write it in the factored
form, because it'll make it easier to simplify later on:
a plus 2 times a plus 2. And now, before we-- let's set
this up like this-- now, before we add the numerators,
it'll probably be a good idea to multiply this out right
there, but let me write the denominator, we know what
that is: it is a plus 2 times a plus 2. Now this numerator: if we have
a minus 2 times a plus 2, we've seen that pattern
before. We can multiply it out if you
like, but we've seen it enough hopefully to recognize that this
is going to be a squared minus 2 squared. This is going to be
a squared minus 4. You can multiply it out, and the
middle terms cancel out-- the negative 2 times a cancels
out the a times 2, and you're just left with a squared minus
4-- that's that over there. And then you have this: you have
minus a minus 3, so let's be very careful here-- you're
subtracting a minus 3, so you want to distribute the negative
sign, or multiply both of these terms
times negative 1. So you could put a minus a here,
and then negative 3 is plus 3, so what does
this simplify to? You have a squared minus a
plus-- let's see, negative 4 plus 3 is negative 1, all
of that over a plus 2 times a plus 2. We could write that as
a plus 2 squared. Now, we might want to factor
this numerator out more, to just make sure it doesn't
contain a common factor with the denominator. The denominator is just 2a
plus 2 is multiplied by themselves. And you can see from inspection
a plus 2 will not be a factor in this top
expression-- if it was, this number right here would be
divisible by 2, it's not divisible by 2. So, a plus 2 is not one of the
factors here, so there's not going to be any more
simplification, even if we were able to factor this thing,
and the numerator out. So we're done. We have simplified the rational
expression, and the domain is for all a's, except
for a cannot, or, all a's given that a does not equal
negative 2-- all a's except for negative 2. And we are done.