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College Algebra
Course: College Algebra > Unit 8
Lesson 5: Adding and subtracting rational expressions- Intro to adding & subtracting rational expressions
- Adding & subtracting rational expressions: like denominators
- Intro to adding rational expressions with unlike denominators
- Adding rational expression: unlike denominators
- Subtracting rational expressions: unlike denominators
- Add & subtract rational expressions (basic)
- Adding & subtracting rational expressions
- Least common multiple of polynomials
- Subtracting rational expressions: factored denominators
- Subtracting rational expressions
- Add & subtract rational expressions
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Subtracting rational expressions: factored denominators
Sal subtracts two rational expressions whose denominators are factored. The denominators aren't the same but they share a factor.
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- can you factor by grouping in the numerator at the end?(8 votes)
- how do I stop getting confused between LCM and GFC(6 votes)
- The *G*reatest*C*ommon*F*actor is based upon what integer divides evenly into two numbers; the *L*east*C*ommon*M*ultiple is based upon what integer two numbers share in a list of multiples.
2. The GCF must be a prime number; the LCM must be a composite number.
Read more: Difference Between GCF and LCM | Difference Between http://www.differencebetween.net/science/difference-between-gcf-and-lcm/#ixzz5mCRcxtx4(6 votes)
- sorry i figured out that letter that i thought was a 2/7 is accually a z(6 votes)
- Shouldn’t you check to see if the final answer can be simplified? Like maybe he could factor by grouping or something like that on the numerator and check to see if there can be anything that can be cancelled out?(3 votes)
- No, because doing that would change the final answer. If, for example, we found a (z+8) factor to cancel, then our final function would be defined at z=-8, while the original is not, so they're different functions.
But as it happens, the polynomial in the numerator has no linear factors anyway, which you can check with the rational root theorem.(3 votes)
- is that number with a line through it a 2 or a 7? Because it looks like a 2 and a 7 mixed together(4 votes)
- Athe inverts the numerator of the second fraction 3:36
3(z+8) becomes -3(z+8). Why is this necessary?
Edit: There is subtraction involved so I imagine this is to accommodate the fact that the signs change when combining them. If this was addition the sign change would not be necessary. Am I wrong?(3 votes)- That is correct. He did invert the coefficient because he wanted to use the addition operator, so he accommodated the fact that one has to change the signs after distributing the subtraction sign.(3 votes)
- How does Sal know not to factor the numerator(I am referring to the final answer). So he would check for common factors to cancel out. Is there a giveaway in expressions that tell you that the numerator and denominator don't have common factors. Or did Sal do this problem beforehand and factored out the numerator and saw that there were no common factors.(2 votes)
- With lots of experience, you may recognize that factoring the numerator would not be helpful. The better approach is to always factor it so you don't miss common factors that can and must be cancelled out.(4 votes)
- How do u find least common denominator with letters and numbers?🤔(1 vote)
- You just treat variable as one of the factors. For example:
70 = 2 x 5 x 7
6b = 2 x 3 x b
So, lcd of 70 and 6b is:2 x 3 x 5 x 7 x b = 210b
(4 votes)
- I still dont get it!! How can you find the LCM if you dont know the denominator in the first place!(2 votes)
- We actually know the denominator, but the only difference is that it is in the form of an algebraic expression rather than integers, so the procedure looks different and complicated. But in fact, it is not(1 vote)
- Why do you not multiply the 3 with (z+8)(9z-5)(z+6) and only with the (z+8) ?(1 vote)
- Each fraction needs to be converted to the common denominator: (z+8)(9z-5)(z+6). The factor with 3 in the numerator already has the factors of (9z-5)(z+6) in the denominator. So, we multiply the fraction by the factor that is missing: (z+8)/(z+8).
If you multiplied by (z+8)(9z-5)(z+6), your denominator would become: (9z-5)(z+6)(z+8)(9z-5)(z+6) which is much bigger than the common denominator that you are trying to create.
Hope this helps.(3 votes)
Video transcript
- [Voiceover] Pause this video and see if you can subtract this magenta rational expression from this yellow one. Alright, now let's do this together. And the first thing that jumps out at you is that you realize these don't have the same denominator and you would like them to have the same denominator. And so you might say,
well, let me rewrite them so that they have a common denominator. And a common denominator that will work will be one that is divisible
by each of these denominators. So it has all the factors of each of these denominators and lucky for us, each of these denominators
are already factored. So let me just write
the common denominator, I'll start rewriting
the yellow expression. So, you have the yellow expression, actually, let me just make it clear, I'm going to write both, the yellow one, and then you're going to
subtract the magenta one. Whoops. I'm saying yellow
but drawing in magenta. So you have the yellow expression which I'm about to rewrite, actually, I'm going to make a longer line, so the yellow expression
minus the magenta one, minus the magenta one, right over there. Now, as I mentioned, we
want to have a denominator that has all, the common
denominator has to have, be divisible by both,
this yellow denominator and this magenta one. So it's got to have
the Z plus eight in it. It's got to have the 9z minus five in it. And it's also got to have both of these. Well, I already, we already
accounted for the 9z minus five. So it has to have, be divisible
by Z plus six. Z plus six. Notice just by multiplying the denominator by Z plus six, we're not
divisible by both of these factors AND both of these factors
because 9z minus five was the factor common to both of them. And if you were just dealing with numbers when you were just adding
or subtracting fractions, it works the exact same way. Alright, so what will
the numerator become? Well, we multiply the denominator times Z plus six, so we have to do the same thing to the numerator. It's going to be negative Z
to the third times Z plus six. Now let's focus over here. We had, well, we want
the same denominator, so we can write this as Z plus eight Z plus eight times Z plus six, times Z plus six times 9z minus five. And these are equivalent. I've just changed the order
that we multiply in it, that doesn't change their value. And if we multiplied the, so
we had a three on top before and if we multiply the
denominator times Z plus eight, we also have to multiply the
numerator times Z plus eight. So there you go. And so, this is going to be equal to, this is going to be equal
to, actually, I'll just make a big line right over here. This is all going to be equal to. We have our, probably
don't need that much space, let me see, maybe that,
maybe about that much. So I'm going to have the same denominator and I'll just write it
in a neutral color now. Z plus eight times 9z minus
five times Z plus six. So over here, just in this blue color, we want to distribute this
negative Z to the third. Negative Z to the third times
Z is negative Z to the fourth. Negative Z to the third times
six is minus 6z to the third. And now this negative
sign, right over here, actually, instead of saying negative Z, negative of this entire thing, we could just say plus
the negative of this. Or another of thinking about it, you could view this as negative
three times Z plus eight. So we could just distribute that. So let's do that. So negative three times Z is negative 3z and negative three times
eight is negative 24. And there you go. We are, we are done. We found a common denominator. And once you have a common denominator, you could just subtract
or add the numerators, and instead of doing this
as minus this entire thing, I viewed it as adding and then having a negative three in the numerator, distributing that and then these, I can't simplify it any further. Sometimes you'll do one of these types of exercises and you might have two second-degree terms
or two first-degree terms or two constants or something like that and then you might want
to add or subtract them to simplify it but here, these
all have different degrees so I can't simplify it any
further and so we are all done.