At 5:32, Sir Sal says that we have to just throw in "another Z" meaning one, but the z factor of 7 has z^2, so shouldn't the final answer be 21z^3 because we are multiplying it.

The LCM contains the highest number of each of the different factors in the given polynomials. The first polynomial has z to the first power; the second polynomial has z to the second power. So only a z^2 is required.

At 3:21 when he reduced everything down, what happened to 2z's z variable? Cause on the third phase of reduction you basically did get 3z(z+1)(z-3) which gives back every variable but not 2z, you only get 2 back.

Remember, when you factor the polynomial, you find factors of -3 that add to -2. The 2 factors are: -3 and +1. So, in factored form, you will not see the -2z. But, if you multiply (z -3)(z+1), you will get the -2z in the middle term. Hope this helps.

When Sal writes down what the LCM should have, shouldn't it be z^3 since it needs z and z^2?

Nope. Z^2 works. Consider the LCM for 2 and 4 (note 4=2^2). The LCM is 4, not 8. When you have multiple instances of the same factor in different denominators, you use the highest exponent for that factor in any *one* denominator. Hope this helps.

shouldn't it(the first part of the expression) be 21z^3 as opposed to ^2 since 3z times 7z^2 equals 21z^3?

You're finding the lowest common multiple. You don't need z^3 in the LCM. Z^2 is enough. If may be easier to understand if you use a number instead of the variable z. Let's say z=2. You are trying to find the lowest common multiple of 2 and 2^2=4. In your approach, you are saying you need to use 2^3=8 as the lowest common multiple. But, that is larger than it needs to be. 4 is a multiple of 2. So, the lowest common multiple is 4 (i.e., 2^2), not 8. The general rule is that the LCM for z vs z^2 is the one with the highest exponent (the z^2) because it is a multiple of the smaller value. Hope this helps.

Why don't you multiply out 21z^2(z+1)(z-3)(z+2)?

You will need the LCM to convert fractions to have a common denominator. The conversion is done by multiplying each fraction with one or more missing factors to make the fraction's denominator match the LCM. Having the LCM in factored form lets you compare the factors already in a denominator with the factors in the LCM to identify the missing factors. If you multiply all the factors together, you can't do this. Hope this helps.

First of all, sorry for my stupidity due to the fact that I'm a high school sophomore, but I was wondering about the least common factor. Why Sal didn't factor 4 as (-2)*(-2) and 6 as (-2)*(-3)? If he did it, what would happen? Will the LCM stay the same or change into (-2)*(-2)*(-3). And is it acceptably correct?

First of all, this isn't a stupid question, it gets at an important feature of how factorization works. Namely, that every integer has a unique prime factorization _up to multiplication by ±1_. ±1 are special integers because their reciprocals are also integers. No other integers have that property. We can say that 6=2·3, or 1·2·3, or (-1)·(-1)·1·2·3, or include any other combination of ±1s. We can't include an extra 3, for example, because we would have to cancel it with a factor of 1/3, and 1/3 isn't an integer. But it's convenient if we can say that every integer has a unique prime factorization, so we standardize what that means by only allowing positive factors and declaring that 1 is not prime. Sal did not write 4=(-2)·(-2) because it would have gone against that standard.

What is the best method to expand (z+1)(z-3)(z+2)? I first expanded (z+1)(z-3) and then multiplied its result by (z+2) to get z^3-7z-6.

Your method is fine. Due to the commutative property of multiplication, the order of the multiplication does not matter. You can pick any 2 of the binomials to multiply first, then multiply the result with the remaining binomial. You would get the same result.

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Algebra (all content)

Course: Algebra (all content) > Unit 13

Lesson 4: Adding & subtracting rational expressions

Least common multiple of polynomials

Name: Least common multiple of polynomials
Uploaded: 2016-04-17T13:29:43Z
Description: Sal finds the least common multiple (LCM) of 3z³-6z²-9z and 7z⁴+21z³+14z².

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Sal finds the least common multiple (LCM) of 3z³-6z²-9z and 7z⁴+21z³+14z².

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Sakshee Pande
Posted 6 years ago. Direct link to Sakshee Pande's post “At 5:32, Sir Sal says tha...”
At
5:32
, Sir Sal says that we have to just throw in "another Z" meaning one, but the z factor of 7 has z^2, so shouldn't the final answer be 21z^3 because we are multiplying it.
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(18 votes)
Answer
- Judith Gibson
  Posted 6 years ago. Direct link to Judith Gibson's post “The LCM contains the high...”
  The LCM contains the highest number of each of the different factors in the given polynomials.
  The first polynomial has z to the first power; the second polynomial has z to the second power.
  So only a z^2 is required.
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  (7 votes)
Rocket Raccoon
Posted 6 years ago. Direct link to Rocket Raccoon's post “At 3:21 when he reduced e...”
At
3:21
when he reduced everything down, what happened to 2z's z variable? Cause on the third phase of reduction you basically did get 3z(z+1)(z-3) which gives back every variable but not 2z, you only get 2 back.
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(4 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Remember, when you factor...”
  Remember, when you factor the polynomial, you find factors of -3 that add to -2. The 2 factors are: -3 and +1. So, in factored form, you will not see the -2z. But, if you multiply (z -3)(z+1), you will get the -2z in the middle term.
  Hope this helps.
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  (4 votes)
blackd
Posted 6 years ago. Direct link to blackd's post “At 6:08, can you FOIL out...”
At
6:08
, can you FOIL out those binomials into a simpler form? Or would you not.

It would become:

21z^2(z^3-7z-6) and then you would further multiply?
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(5 votes)
Answer
- khushi2014
  Posted 5 years ago. Direct link to khushi2014's post “You can if you prefer...i...”
  You can if you prefer...it would not change the answer, just undo its simplification.
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  (1 vote)
Minh
Posted 3 years ago. Direct link to Minh's post “First of all, sorry for m...”
First of all, sorry for my stupidity due to the fact that I'm a high school sophomore, but I was wondering about the least common factor. Why Sal didn't factor 4 as (-2)*(-2) and 6 as (-2)*(-3)? If he did it, what would happen? Will the LCM stay the same or change into (-2)*(-2)*(-3). And is it acceptably correct?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “First of all, this isn't ...”
  First of all, this isn't a stupid question, it gets at an important feature of how factorization works. Namely, that every integer has a unique prime factorization up to multiplication by ±1.
  
  ±1 are special integers because their reciprocals are also integers. No other integers have that property. We can say that 6=2·3, or 1·2·3, or (-1)·(-1)·1·2·3, or include any other combination of ±1s. We can't include an extra 3, for example, because we would have to cancel it with a factor of 1/3, and 1/3 isn't an integer.
  
  But it's convenient if we can say that every integer has a unique prime factorization, so we standardize what that means by only allowing positive factors and declaring that 1 is not prime. Sal did not write 4=(-2)·(-2) because it would have gone against that standard.
  Comment on kubleeka's post “First of all, this isn't ...”
  (9 votes)
masterchief283
Posted 4 years ago. Direct link to masterchief283's post “How do you use this for m...”
How do you use this for monomials?
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(3 votes)
Answer
Nolan Duarte
Posted 2 months ago. Direct link to Nolan Duarte's post “When Sal writes down what...”
When Sal writes down what the LCM should have, shouldn't it be z^3 since it needs z and z^2?
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(2 votes)
Answer
- Kim Seidel
  Posted 2 months ago. Direct link to Kim Seidel's post “Nope. Z^2 works. Conside...”
  Nope. Z^2 works.
  Consider the LCM for 2 and 4 (note 4=2^2). The LCM is 4, not 8.
  
  When you have multiple instances of the same factor in different denominators, you use the highest exponent for that factor in any one denominator.
  
  Hope this helps.
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  (4 votes)
Amy Wang
Posted 4 years ago. Direct link to Amy Wang's post “shouldn't it(the first pa...”
shouldn't it(the first part of the expression) be 21z^3 as opposed to ^2 since 3z times 7z^2 equals 21z^3?
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(2 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “You're finding the lowest...”
  You're finding the lowest common multiple. You don't need z^3 in the LCM. Z^2 is enough. If may be easier to understand if you use a number instead of the variable z. Let's say z=2.
  You are trying to find the lowest common multiple of 2 and 2^2=4. In your approach, you are saying you need to use 2^3=8 as the lowest common multiple. But, that is larger than it needs to be. 4 is a multiple of 2. So, the lowest common multiple is 4 (i.e., 2^2), not 8.
  
  The general rule is that the LCM for z vs z^2 is the one with the highest exponent (the z^2) because it is a multiple of the smaller value.
  
  Hope this helps.
  Comment on Kim Seidel's post “You're finding the lowest...”
  (3 votes)
Ron Shlomi
Posted 5 years ago. Direct link to Ron Shlomi's post “Why don't you multiply ou...”
Why don't you multiply out 21z^2(z+1)(z-3)(z+2)?
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(2 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “You will need the LCM to ...”
  You will need the LCM to convert fractions to have a common denominator. The conversion is done by multiplying each fraction with one or more missing factors to make the fraction's denominator match the LCM. Having the LCM in factored form lets you compare the factors already in a denominator with the factors in the LCM to identify the missing factors. If you multiply all the factors together, you can't do this.
  
  Hope this helps.
  Button navigates to signup page
  (3 votes)
ANB
Posted a year ago. Direct link to ANB's post “What is the best method t...”
What is the best method to expand (z+1)(z-3)(z+2)? I first expanded (z+1)(z-3) and then multiplied its result by (z+2) to get z^3-7z-6.
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(1 vote)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Your method is fine. Due...”
  Your method is fine. Due to the commutative property of multiplication, the order of the multiplication does not matter. You can pick any 2 of the binomials to multiply first, then multiply the result with the remaining binomial. You would get the same result.
  Comment on Kim Seidel's post “Your method is fine. Due...”
  (3 votes)
Evan
Posted 4 years ago. Direct link to Evan's post “At 5:12 if Sal had a 6z^2...”
At
5:12
if Sal had a 6z^2 (instead of a 7z^2) would he still have to multiply the bottom expression by 6, or could he just change the pink 3 to a 6 (since 6 has both 3 and 6 as factors)

Hope this makes sense. :)
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(2 votes)
Answer

Video transcript

- [Voiceover] So they're asking us to find the least common multiple of these two different polynomials. So the first one's three z to the third minus six z squared minus nine z and the second is seven z to the fourth plus 21 z to the third plus 14 z squared. Now, if you're saying, "Well, what is a," you're familiar with least common multiple of two numbers. One way to think about them is if I were to find, say, the least common multiple between, I don't know, four and six, you literally could look at all the multiples and see which one is least. So you go four, eight, 12, 16, so on and forth. You're gonna do the same thing for six. You'd go six, 12, 18, 24, so on and so forth. And you immediately see that they do have, they'll actually have multiple common multiples, but the least of the common multiples you immediately see is going to be 12. Now another way to think about it is to actually factor these numbers out. We could view four, you could view four as being two times two if we look at its prime factorization. And six is two times three. So you could say that the least common multiple, the LCM, of four and six is going to be, well, it's gonna be equal to, we're gonna have to have the factors of both of them. So it's gonna have to have two fours, two times two. It's gonna have to have a two and a three. Well we already have a two. In fact, we have to them. So in order to be divisible by three, we have to be, in order to be divisible by six, we have to have three as one of the factors. And so when you look at it that way, you say, "Hey look, we have to contain "all of the factors of each," we have to have at least two twos and we have to have at least one three because the two twos take care of this one too right over there. And you see that this is also going to be equal to 12. Now when we think about it for polynomials, we're gonna think about it a little bit more. It's essentially the same idea, but we're gonna think about it a little bit more with the second lens. We're gonna think about the factors and say, well the least common multiple needs to contain the factors of both but it shouldn't contain more than that. You can always find a multiple of two polynomials by just multiplying them, but we don't wanna find just any multiple. We wanna find the least common multiple. So let's factor each of them. So this first one, three z to the third minus six z squared minus nine z. Let's see. Immediately, let's see, all of these terms are divisible by three z, so let's factor out of three z. So it's three z times z squared. We factor out of three z out of that. See, it's gonna be minus two z if we factor out of three z out of that. And then minus three. And notice, if you were to distribute this three z back, you would get exactly what we have up here. And so let's see, can we factor this further? Can we think of two numbers that if we multiply, we get negative three and if we add them, we get negative two? And one's gonna be positive, one's gonna be negative since the product is negative. So let's see. It sounds like negative three and positive one. So we can rewrite this as three z times z plus one times z minus three. I think I have factored this first polynomial about as much as I can. One times negative three is negative three. One z minus three z is negative two z, so that looks good. So now let's factor, now let's factor this other character over here, this fourth degree polynomial. So, every one of those terms look like they're visible by seven z squared, so I could write this as seven z squared times z squared. If we factor out of seven z squared here, you're just left with z squared. And then plus, 21 divided by seven is three, z to the third divided by z squared is z. And then plus, 14 divided by seven is two. z squared divided by z squared is one. So it's just gonna be a two there. And so this is gonna be the same thing as seven z squared. And this can be factored into, let's see, two times one is two, two plus one is three. So, z plus one times z plus two. Now let's think about the least common multiple. We've factored each of these just the way that we, when we did the prime factorization for regular numbers. Now we have factored this down to simpler expressions as we will find useful. And so, the least common multiple of these two things has to contain each of these factors. So the least common multiples got to contain a three z. It's got to contain, and let me expand it out a little bit. It's got to contain a three. It's got to contain a z. It's gotta contain a z plus one, z plus one. I don't have to write the little dot there. It's got to contain a z plus one. It's gotta contain a z minus three. Let's see. It's gotta contain a seven. We do not have a seven here yet, so we have to include a seven, so I'll put the seven out front with the numbers. It's gotta include a seven. It's gotta include a z squared. Well we only have a z right now, so let's throw in another z. So I could throw in another, I could write, I could put a z out front or I could just make this a squared. It still contains that z, but now we contain another z, or multiplying by another z to have z squared. See, we already have a z plus one in here. We need a z plus two as well, z plus two as well. And there you have it. This is the least common multiple. If I were to write it all out in a neutral color, it's gonna be 21 z squared times z plus one times z minus three times z plus two (chuckle) I say two and I write six. z plus two. And we are all done. And I really want you to appreciate. This is the exact same thing we're doing when we're doing, a very similar thing than what we're doing when we're finding these common multiples of regular numbers, looking at their factors. In the case of numbers, prime factors. And then we say, "Okay. The least common multiple "has to be a superset, has to contain all of these, "but we don't wanna contain," you know, I could multiply this times a hundred, it's still gonna be a common multiple of these two, but it's no longer the least common multiple. Likewise, 12 is the least common multiple of four and six. If I just wanted a common multiple, I can multiply that times a hundred, 1200 would also be a multiple of four and six, but it wouldn't be the least common multiple, so we don't wanna do that. Hopefully you found that interesting.