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# Subtracting rational expressions

CCSS.Math:

## Video transcript

find the difference express the answer as a simplified rational expression and state the domain so we have two rational expressions and we're subtracting one from the other and just like when we first learn to subtract fractions or add fractions we have to find a common denominator and the best way to find a common denominator for 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's divisible by the two denominators here so this guy right here is completely factored he's just a plus two this one over here let's see if we can factor it a squared plus 4a plus four well you see the pattern the four 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 so we say it's a plus 2 times a plus 2 that's what a squared plus 4a plus 4 is so this is obviously divisible by itself everything is divisible by itself except I guess 4 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 so let's set that up so let's set them this will be the same thing as being equal to this first term right here a minus 2 over a plus 2 a minus 2 over a plus 2 but we want the denominator now to be a plus 2 times a plus 2 we want it to be a plus 2 squared so let's multiply this numerator and denominator by a plus 2 so it's denominator is the same thing as this so let's multiply both the numerator and the denominator by a plus 2 by a plus 2 and we're going to assume that a is not equal to negative 2 that would have made this undefined it would have also made this undefined so 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 this denominator is already the common denominator so 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 in the factored form 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 make 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 or this is going to be a squared minus 4 you can multiply it out the middle terms cancel out the negative 2 times a cancels out with 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 - 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 negative 3 is plus 3 so what does this simplify to you have a squared a squared minus a minus a plus let's see negative 4 plus 3 is negative 1 all of that over all of that over a plus 2 times a plus 2 or we could just rewrite we could write that as a plus 2 squared 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 2 a plus 2 is multiplied by themselves and you can see from inspection a plus 2 will not be a factor in this top expression I mean 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 wanted you know even if we even if we were able to factor this thing in the numerator out so we're done we have simplified the rational expression and the domain is all for all A's except for a cannot or all A's given that a does not equal negative to all A's except for negative two and we are done