Find LCM of Polynomials...2

[feliz_nyc]

College Algebra
Chapter 1/Section 7

 

[Dr.Peterson]

The first looks good.


Does what you wrote fit the definition? Does it if you insert an exponent of 2?

 

[Steven G]

You can check for yourself.

Does x^2+4x+4 = (x+2)^2 go evenly into x^2(x+2)?
Does x^3+2x^2 = x^2(x+2) go evenly into x^2(x+2)?
Does (x+2)^3 go evenly into x^2(x+2)?

If you say yes to all three questions above, then you DO have a common multiple. It may NOT be the least common multiple but you can still use it.
If you say no to any of the three questions below, then you DO NOT have a common multiple. Try again.

Multiplying by the LCM or CM is meant to kill off/absorb ALL the denominators

 

[feliz_nyc]

You can check for yourself.

Does x^2+4x+4 = (x+2)^2 go evenly into x^2(x+2)?
Does x^3+2x^2 = x^2(x+2) go evenly into x^2(x+2)?
Does (x+2)^3 go evenly into x^2(x+2)?

If you say yes to all three questions above, then you DO have a common multiple. It may NOT be the least common multiple but you can still use it.
If you say no to any of the three questions below, then you DO NOT have a common multiple. Try again.

Multiplying by the LCM or CM is meant to kill off/absorb ALL the denominators

The first looks good.

View attachment 33624
Does what you wrote fit the definition? Does it if you insert an exponent of 2?

How about now?

 

[Dr.Peterson]

How about now?

That is a common multiple, since all the factors of each expression are in it.

But it is not the LEAST common multiple, because there are factors in it that are not needed. No one expression has (x-3) as a factor TWICE, so you don't need the square. The same is true of (x+3).

One way I like to think about the LCM is to start with one expression (in factored form), usually the biggest one, and then ask what more I need in order to be a multiple of the others. If we start with x(x-3)(x+3), we find that all of x(x+3) is already there, so nothing needs to be added; and the same is true of (x-3). So in fact the third expression you were given, being already a multiple of each of the others, is itself the LCM.

And a way to check it is to go through each distinct factor, and ask myself whether I have exactly as many of it as I need. For example, the first factor is x; it does not occur in (x-3), and occurs in each of x(x+3) and x(x-3)(x+3) exactly once, so the exponent, 1, is correct. And so on.

Try the same thinking with the second problem. Is it a common multiple? Is it the least common multiple?

 

[feliz_nyc]

View attachment 33632

That is a common multiple, since all the factors of each expression are in it.

But it is not the LEAST common multiple, because there are factors in it that are not needed. No one expression has (x-3) as a factor TWICE, so you don't need the square. The same is true of (x+3).

One way I like to think about the LCM is to start with one expression (in factored form), usually the biggest one, and then ask what more I need in order to be a multiple of the others. If we start with x(x-3)(x+3), we find that all of x(x+3) is already there, so nothing needs to be added; and the same is true of (x-3). So in fact the third expression you were given, being already a multiple of each of the others, is itself the LCM.

And a way to check it is to go through each distinct factor, and ask myself whether I have exactly as many of it as I need. For example, the first factor is x; it does not occur in (x-3), and occurs in each of x(x+3) and x(x-3)(x+3) exactly once, so the exponent, 1, is correct. And so on.

Try the same thinking with the second problem. Is it a common multiple? Is it the least common multiple?

Are you saying that both LCM answers are wrong? If that is the case, I will practice some more by watching YouTube clips. I am sure this is no big deal. For some reason, I haven't caught on to the simple idea.

 

[Dr.Peterson]

Are you saying that both LCM answers are wrong? If that is the case, I will practice some more by watching YouTube clips. I am sure this is no big deal. For some reason, I haven't caught on to the simple idea.

I didn't say the second was wrong; I said it will be good practice to check it. But you do need to fix the first one and show it to us.

I'm not sure that watching videos is the same thing as practicing; seeing is not doing. What you need is to see an orderly way to (a) know what the LCM is, and (b) find it; and then to practice doing that. I gave you one approach, but I haven't seen that you've tried it.

There's a more standard method at the end of this page, which introduces it using the LCM of numbers.

I've been looking for good sources, and have found none to recommend.

 

[feliz_nyc]

I didn't say the second was wrong; I said it will be good practice to check it. But you do need to fix the first one and show it to us.

I'm not sure that watching videos is the same thing as practicing; seeing is not doing. What you need is to see an orderly way to (a) know what the LCM is, and (b) find it; and then to practice doing that. I gave you one approach, but I haven't seen that you've tried it.

There's a more standard method at the end of this page, which introduces it using the LCM of numbers.

I've been looking for good sources, and have found none to recommend.

My answer for the first problem is x(x-3)(x+3) which turns out to be x(x^2 - 9) when multiplied.
You say?

By the way, I used the following as my guide:

 

[feliz_nyc]

View attachment 33632

That is a common multiple, since all the factors of each expression are in it.

But it is not the LEAST common multiple, because there are factors in it that are not needed. No one expression has (x-3) as a factor TWICE, so you don't need the square. The same is true of (x+3).

One way I like to think about the LCM is to start with one expression (in factored form), usually the biggest one, and then ask what more I need in order to be a multiple of the others. If we start with x(x-3)(x+3), we find that all of x(x+3) is already there, so nothing needs to be added; and the same is true of (x-3). So in fact the third expression you were given, being already a multiple of each of the others, is itself the LCM.

And a way to check it is to go through each distinct factor, and ask myself whether I have exactly as many of it as I need. For example, the first factor is x; it does not occur in (x-3), and occurs in each of x(x+3) and x(x-3)(x+3) exactly once, so the exponent, 1, is correct. And so on.

Try the same thinking with the second problem. Is it a common multiple? Is it the least common multiple?

Here is the book I will use moving forward:

 

[feliz_nyc]

I didn't say the second was wrong; I said it will be good practice to check it. But you do need to fix the first one and show it to us.

I'm not sure that watching videos is the same thing as practicing; seeing is not doing. What you need is to see an orderly way to (a) know what the LCM is, and (b) find it; and then to practice doing that. I gave you one approach, but I haven't seen that you've tried it.

There's a more standard method at the end of this page, which introduces it using the LCM of numbers.

I've been looking for good sources, and have found none to recommend.

Disregard the college algebra textbook I reported to be using. After downloading the entire book, I noticed that the words are not clear, a bit fuzzy. I will download another textbook.

 

[Steven G]

My answer for the first problem is x(x-3)(x+3) which turns out to be x(x^2 - 9) when multiplied.

Incomplete. When you multiply out x(x-3)(x+3) you get x(x-3)(x+3) = x(x^2 - 9) = x^3-9x.
My question is, why do you want to multiply out?

 

[feliz_nyc]

Incomplete. When you multiply out x(x-3)(x+3) you get x(x-3)(x+3) = x(x^2 - 9) = x^3-9x.
My question is, why do you want to multiply out?

We need to multiply the final result to show the LCM in simplified form. Yes?

 

[topsquark]

We need to multiply the final result to show the LCM in simplified form. Yes?

That depends on what you need the form for. Usually if you are looking for an LCM you would want it to be factored. I'd leave it that way unless there is a specific reason not to.

-Dan

 

[Steven G]

We need to multiply the final result to show the LCM in simplified form. Yes?

What final result are you referring to?
What do you mean when you say we need to multiply the final result to show the LCM in simplified form?

 

[feliz_nyc]

What final result are you referring to?
What do you mean when you say we need to multiply the final result to show the LCM in simplified form?

For example, if the LCM is (x + 2)(x + 3), it is best to FOIL out the two factors to become x^2 + 5x + 6 as a final answer. This is just an example.

 

[feliz_nyc]

That depends on what you need the form for. Usually if you are looking for an LCM you would want it to be factored. I'd leave it that way unless there is a specific reason not to.

-Dan

I am happy to be a member of this site. It is my hope to learn as much as I can. What is the ultimate goal here at age 57? My goal is to learn Calculus l, ll, and lll. I have seen video clips of calculus and find the course to be super interesting. You say?

 

[Dr.Peterson]

For example, if the LCM is (x + 2)(x + 3), it is best to FOIL out the two factors to become x^2 + 5x + 6 as a final answer. This is just an example.

As you were told in #13, this is not true:

That depends on what you need the form for. Usually if you are looking for an LCM you would want it to be factored. I'd leave it that way unless there is a specific reason not to.

-Dan

In general, because factoring is hard and "foiling" is easy, we prefer to leave an expression in factored form once we have it like that, unless there is reason to prefer another form. The main reason to prefer the expanded form is when the next thing you want to do is to add two of them. When you don't know what will be done with your answer, you leave it as is.

It is my hope to learn as much as I can.

If you want to learn, pay attention to what experienced people tell you. Don't just ignore them and insist that you're right.

 

[topsquark]

I am happy to be a member of this site. It is my hope to learn as much as I can. What is the ultimate goal here at age 57? My goal is to learn Calculus l, ll, and lll. I have seen video clips of calculus and find the course to be super interesting. You say?

I think video clips are useful. I think textbooks are necessary. I think a class at a College is by far the best.

And, of course, we're happy to help where we can.

Best of luck!

-Dan

 

[feliz_nyc]

As you were told in #13, this is not true:

In general, because factoring is hard and "foiling" is easy, we prefer to leave an expression in factored form once we have it like that, unless there is reason to prefer another form. The main reason to prefer the expanded form is when the next thing you want to do is to add two of them. When you don't know what will be done with your answer, you leave it as is.

If you want to learn, pay attention to what experienced people tell you. Don't just ignore them and insist that you're right.

Copy. By the way, I do pay attention to detail and always do my best to learn from others who are more experienced and/or have a math degree. Let's be honest here.

I am 57. My college days ended in 1994. I am not going to be a math teacher. Perhaps a tutor to my nephews and nieces who are about to start middle school and high school. Math is like crossword puzzles. It helps my brain cells to stay active and alive. No pun intended.

 

[feliz_nyc]

I think video clips are useful. I think textbooks are necessary. I think a class at a College is by far the best.

And, of course, we're happy to help where we can.

Best of luck!

-Dan

Thank you. Realistically speaking, at age 57 returning to college is not an option. By the way, I took precalculus in the Spring 1993 semester and earned an A minus. Not bad for someone who did not major in mathematics. You see, I can learn math concepts when people are patient with my learning style and not critical of my constant errors.