Algebra II (2018 edition)
- Multiplying & dividing rational expressions: monomials
- Multiplying rational expressions
- Dividing rational expressions
- Multiply & divide rational expressions: Error analysis
- Multiplying rational expressions
- Dividing rational expressions
- Multiply & divide rational expressions
- Multiplying rational expressions: multiple variables
- Dividing rational expressions: unknown expression
- Multiply & divide rational expressions (advanced)
Sal finds the polynomial D for which (20y²-80)/D ÷ (4y²-8y)/(y³+9y²)=1 is true for all values of y for which it's defined.
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- how did 20(y2 + 4) become 20 (y+2) (y-2)(7 votes)
- they factored a difference of squares. To see that they are equivalent, multiply (x+2) and (x-2). These factors are "conjugates" of one another so the middle "x" terms cancel out and you are left with just the x^2 term and the constant 4 term. This video can provide more background if you need it.
- Does this mean that D cannot be equal to zero? If D were to be equal to zero, the first fraction would be undefined.(6 votes)
- You're right. If for any a, b, c that is a real number and d is 0, (a/b)/(c/d) would be undefined; so the entire result would be undefined. Hence, D can't be equal to 0.(7 votes)
- At about3:25Sal distributes D across only 4y but doesn't put parenthesis around 4y(y - 2) so that D can be multiplied across the whole expression. Why doesn't he do that?(2 votes)
- Sal doesn't want to distribute the D. He just wants to multiply it with the expression. You only distribute if you are adding or subtracting 2 terms. So if the problem were 4y + (y-2) Then he would distribute the D. Essentially whether you distribute in multiplication or not you will get the same answer.(2 votes)
- what can you do when you cant multiply the reciprocal(2 votes)
- What if you cannot get the variable "D" by itself by cancelling?(2 votes)
- As long as D is by itself int he denomnator (or numerator) or just has some other term being multiplied by it you can always get D by itself.
If you had something like D-1 or D + x or something in the denominator you would want to get that whole term by itself, so get everything else = D+x then just subtract the x, or whatever everything else happens to be. It might be a little messier, but that will be the answer.
If you like I cans show this with an example or two.(1 vote)
- I figured out how to do this in an alternative way or think about it. I just multiplied both sides by D to get rid of the D on the left hand side to put it on the ride hand side then follow the rules on rational expression division to deal with the left hand side since the D has been cancelled out.(2 votes)
- Hopefully, you multiplied both sides by D. If you divided by D, it doesn't cancel out the D. You would end up with D^2 in the denominator.
If you did use multiplication, then your technique does work.(1 vote)
- On my worksheet, we have to Multiply or Divide. Simplify the answer if needed.
This is the question that I have been stuck on, and I don't know what to do now, could someone please help!
(15-5x) / (x²-x-6) ÷ (5x) / (x²+6x+8)?
I know that you would have to factor, which gives me:
-5(x-3) / (x+2)(x-3) ÷ (x+4)(x+2) / (5x)
But when I cancel stuff out, I get:
-5 times (x+4) / (5x)
What do I do?(1 vote)
- You have factored everything correctly. However, when you flip the numerator and denominator, you have to change the division symbol to a multiplication symbol. Perhaps that's what you're stumbling on. If you still have any questions, don't hesitate to ask! I'd be glad to help :)(2 votes)
- Two questions:
1. On the Unit Test for Multiplying And Dividing Rational Expressions, the factored form of problem #5 was D= 5y(y+2)(y+9). I multiplied the polynomial by multiplying the two terms in parenthesis and then multiplied the product by 5y. However, when I got a different answer when I multiplied 5y by the 1st term in parenthesis and then multiplied that product by the term in the 2nd parenthesis. I thought it did not matter which order I multiplied the all 3 factors, for example I thought a*b*c was the same as multiplying b*c*a. Was I wrong because I did not multiply in order?
2. When I finished Unit Test 2 I didn't get any Mastery Points and I got all 5 questions correct the first time.(1 vote)
- Yes, a*b*c is the same as b*c*a
Without seeing your work, it is hard to say what you did wrong. But somewhere there is an error. Comment back with your work and I'll help you find it.(1 vote)
- would 5(y+2)y(y+9)=D also be correct?(1 vote)
- Yes, that would be ok. Commutative property of multiplication applies so the order of the factors doesn't matter. Though it is convention to put the 5y together up front.(1 vote)
- [Voiceover] We're told the following equation is true for all real values of y for which the expression on the left is defined and D is a polynomial expression. And they have this equation here. What is D? All right. So essentially, what they're saying is they don't want us to somehow solve this equation. They're saying D is going to be some type of a polynomial expression. They tell us that right over there: D is a polynomial expression. And if we figure out what D is, this left-hand side of the expression is going to evaluate to one for all real values of y for which the expression is defined. So let's think about how we would tackle that. Well the first thing that pops into my mind, if I'm dividing by a fraction or a rational expression, that's the same thing as multiplying by the reciprocal. So let's just rewrite this on the left-hand side. So this is 20 y squared minus 80 over D times four, oh, let me do the reciprocal, let me be careful, times the reciprocal of this. If I divide by something, the same thing as multiplying by the reciprocal, so let me just swap the numerator and the denominator, (chuckle) numerators and denominators, numerator and denominator. All right, y to the third plus nine y squared, all of that over four y squared minus eight y. That's going to be equal to one. Now let's say if we can simplify all of these business on the left-hand side a little bit. So let's see. Over here, I can divide both terms by 20. So let me factor out 20 because I think it's gonna end up being a difference of squares. So if I factor out of 20, so this is about the same thing as 20 times y squared minus four and y squared minus four, we can rewrite as y plus two times y minus two. It is a difference of squares. So let me write y plus two times y minus two. All right. This down here, four y squared minus eight y, well it looks like we can factor out of four y. And so this is going to be same thing as four y times y minus two. All right. So let me cross that out. So it's the same thing as four y times y and minus two and I already see that this y minus two here and this y minus two here are gonna cancel out. And let's see. Up here, both terms are divisible by y squared, so I can rewrite this as, I don't know if this is actually going to be helpful because it's gonna, well let me just do it, just in case. So that's the same thing as y squared times, y squared times y plus nine. All right. And so we can rewrite all of these things. If we were to multiply everything together, we would end up getting, the numerator would get 20 times y plus two times y minus two times y squared times y plus nine, I'm just multiplying all the numerators, and that's going to be over, in the denominator, I would have whatever the expression D is, times four y times y minus two, and that's all going to be equal to one. Now let's think about it. We can divide, we have y minus two divided by y minus two, so those cancel out. Now let's see. We have a... We can divide the numerator and the denominator by y, so that would just become one. And then that would just become a y to the first power. And so, what we'd be left with, what we'd be left with in the numerator is 20 times this y times y plus two times y plus nine over, over four D is equal to one. Now if we wanna solve four D, well we could just multiply both sides by D and one times D is just gonna be D, so we're gonna have D equals something over here and we'd be done. So let's do that. D times that and let's multiply that times D. Let me be clear what I'm doing here. Let me draw a little divider here to make it clear that that's happening in the other side of the page. All right. So D times this, those cancel out, and we're gonna be left with 20 y times, actually, let me, I could simplify it even more. 20 divided by four is five. So, our numerator is now just one. So we have five times y times y plus two times y plus nine is equal to D, and we're done. This is D. This is the polynomial, whoops, that is... That is the polynomial expression that we are looking for. If you were to substitute this back in and then try to simplify it, well, you would end up with all of these over here and D would be this, and so, it would all just cancel out and you would be left with one. For all real values y for which the expression on the left is actually defined. There are some values of y for which the expression on the left is not defined. If y is equal to zero, this denominator is zero and you're dividing by zero, well that's not defined. And then when you multiply by the reciprocal, if this were to become zero, then that wouldn't be cool either. And there's multiple ways to make this equal to zero. y could be equal to negative nine, that would also make this bottom zero. So we could think about that if we wanted to, but they're not asking us to do that. They're saying, for all of the real values for y for which the expression defined, find the D that makes all of this business equal to one and we just did that.