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Adding and Subtracting Rational Expressions  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Whenever we are adding and subtracting rational expressions the first thing we need to do is get a common denominator. And the easiest way to typically do that is to factor out all the denominators we have.
So looking at this x minus 2 nothing I can do with that x² plus 2x plus 4 that doesn’t factor either, but what we end up at the end is x³ minus 8. Hopefully you recognize that as the difference if cubes so we can factor that down to x minus 2, x² plus 2x plus 4 which can be neatly enough is the two denominators that we already have. So x³ minus 8 or this factored version is our least common denominator.
So to make our denominators be all the same I need to multiply this first term by x² plus 2x plus 4 over the same thing. Remember you always have to multiply everything by 1 to keep everything the same, and we need to multiply this second one by x minus 2 over x minus 2. Our third statement already has our common denominator.
So what I’m going to do now is strictly focus on the numerator. I know all my denominators are the same, they all are basically x³ minus 8, and so I’m going to focus on the numerator. Some teachers will want you to do this some will want you to write that denominator out throughout the entire thing. The important thing that you keep in mind is that you are focusing just on the numerator and whatever we are doing is really over x³ minus 8 or same thing as x minus 2, x² plus 2x plus 4.
So what we have to do is multiply this out. So what I end up with is 2x² plus 4x plus 8 minus, and then for this one make sure we remember to distribute this minus sign through. So I’m not to leave it out for right now we have minus x² minus 2x and then minus 3. Distributing this minus sign through, 2x² plus 4x plus 8 minus x² plus 2x minus 3, and then combining like terms. So 2x² and x² leave us with x² so we use those up. We have plus 4x and plus 2x plus 6x plus 8 and minus 3 which is going to be plus 5.
So what I have right now is just the numerator this can actually be factored is we wanted to, it's going to be x plus 1 times x plus 5, and this is where students frequently mess up. They frequently box this in and call this their answer. But remember I was just focusing strictly on the numerator. I had my denominator up here that I sort of left off to the side to save writing. So I need to make sure I throw that back in as well.
So we have this over you could either write it as these two things multiplied together or you could just write it as the x³ minus 8, it doesn’t really matter because they are really the same thing.
Sometimes when we finish up we’ll be able to cancel things so you know if this was an x plus 2 I’d be able to cancel those off, in this case nothing cancels and what I have is just my end product.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Sample Problems (9)
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