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### Course: Integrated math 3>Unit 13

Lesson 8: Adding subtracting rational expressions (factored)

# Least common multiple of polynomials

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

## Want to join the conversation?

• At , 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 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.
• At , 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?
• You can if you prefer...it would not change the answer, just undo its simplification.
(1 vote)
• 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.
• 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.
• How do you use this for monomials?
• 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.
• 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.
(1 vote)
• 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.
• At 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. :)