# Homework Help: Polynomials divisible by

1. Sep 25, 2009

### um0123

Polynomials divisible by....

1. The problem statement, all variables and given/known data

1) Explain why

$$n^3 - n$$

is always divisible by 3 for any n that is an element of the natural numbers.

2) Give 2 other polynomials that are always divisble by 3.

3) Give 2 polynomials that are divisible by 2 but not 4, and 2 other polynomials that are divisible by 2 and 4.

2. Relevant equations

$$n^3 - n$$

3. The attempt at a solution

Im just really confused and dont know where to start with this one. I just need some hints to finding out why $$n^3 - n$$ is divislble by three. And some help for finding a method for finding polynomials divisible by a specific number (if such a method exists).

2. Sep 25, 2009

### Dick

Re: Polynomials divisible by....

In this case n^3-n=(n-1)*n*(n+1). What does that tell you? Can you generalize that to find a polynomial divisible by any number?

3. Sep 25, 2009

### um0123

Re: Polynomials divisible by....

the factored form $$n(n-1)(n+1)$$ tells me that there are zeros at 1 and -1 and 0. But i dont understand what you mean by "generalize that to find a polynomial divisible by any number."

4. Sep 26, 2009

### nietzsche

Re: Polynomials divisible by....

what happens when you multiply any three consecutive integers? what properties will the product have?

5. Sep 26, 2009

### Dick

Re: Polynomials divisible by....

We aren't talking about where the zeros are, we are talking about divisibility by 3. Another way to check is to evaluate n^3-n mod 3 for n=0, 1 and 2. Have you done stuff like that?

6. Sep 26, 2009

### um0123

Re: Polynomials divisible by....

we havent learned modulo in class, but i know what it is from my experience with c++. I dont know if he expects us to use it, but as long as i solve the problem in a manner that he sees how i got my answer it is fine if i do that.

if i check the mod of 3 for 0,1, and 2, it doesnt prove for when n goes to infinity though.

7. Sep 26, 2009

### slider142

Re: Polynomials divisible by....

You don't really need the modulus argument. Can you find three consecutive integers that are not divisible by 3? If so, what are they? If not, why not? As a hint, note that 3 consecutive integers can always be written in the form n, n+1 and n+2 for some n. Can you show what happens if none of them are divisible by 3? (This is a lead to a proof by contradiction argument)

8. Sep 26, 2009

### um0123

Re: Polynomials divisible by....

i dont understand why n, n+1, and n+2 is involved. Is that the same as n, n-1, n+1? but how do i logically prove that its always divisible by three up to inifinity, and how am i supposed to come up with my own polynomials?

9. Sep 26, 2009

### nietzsche

Re: Polynomials divisible by....

if you have three consecutive integers, can you always factor out a three? yes, because if there are three consecutive integers, one of them has to be a multiple of 3, because you're choosing three consecutive integers.

10. Sep 26, 2009

### um0123

Re: Polynomials divisible by....

but you cant factor out a 3 from all of them, only the one divisible by three, so i am confused on how the entire thing is able to factor out a three.

11. Sep 26, 2009

### slider142

Re: Polynomials divisible by....

?? If b is divisible by 5, then abc is also divisible by 5 irrespective of whether a or c are divisible by 5.

12. Sep 26, 2009

### um0123

Re: Polynomials divisible by....

i dont doubt you, but i dont see how that works.

13. Sep 26, 2009

### nietzsche

Re: Polynomials divisible by....

well...take for example 4, 5, 6. you can write 6 as 3x2.

so 4x5x6 = 4x5x3x2 and if you divide that by 3, you get 4x5x2.

you can write any number that is divisible by 3 as 3k, where k is an integer.

14. Sep 26, 2009

### um0123

Re: Polynomials divisible by....

OH, i get it now. I was just being really dumb. okay, so to make polynomials divisible by a specific number all i need to do is have part of it divisible by that number.

i wanna try this now, but i gotta go......:(

15. Sep 26, 2009

### slider142

Re: Polynomials divisible by....

Straightforward. If b is divisible by 5, then b = 5*n for some n. Likewise, abc = a*5*n*c = 5*a*n*c so abc is divisible by 5.

16. Sep 27, 2009

### um0123

Re: Polynomials divisible by....

im having trouble coming up with polynomials that are divisible by 2 but not 4.

Last edited: Sep 27, 2009
17. Sep 27, 2009

### Bohrok

Re: Polynomials divisible by....

What are the terms in the sequence of numbers divisible by 2 but not 4?

18. Sep 27, 2009

### um0123

Re: Polynomials divisible by....

2, 6, 10, 14....

but i tried to put them in and it didnt work out.

19. Sep 27, 2009

### Dick

Re: Polynomials divisible by....

Put them into what? Can't you find a polynomial p(x) such that p(1)=2, p(2)=6, p(3)=10 etc?

20. Sep 27, 2009

### Bohrok

Re: Polynomials divisible by....

Try graphing those points and find a linear equation that fits those points.

21. Sep 28, 2009

### um0123

Re: Polynomials divisible by....

got it, thanks for the help everyone!

22. Sep 28, 2009

### Staff: Mentor

Re: Polynomials divisible by....

An integer that is divisible by 2 but not by 4 has only a single factor of 2 in its factorization. An integer that is divisible by 4 has at least two factors of 2 in it factorization. Can you come up with a polynomial that has only one factor of 2?