# Math Help - factoring of polynomials...

1. ## factoring of polynomials...

I havn't done too much factoring, so what I'm wondering is if there is way to get rid of the exponentials in the last two equations?

n^2-n = n(n-1)
n^3-n = n(n+1)(n-1)
n^4-n = n(n^3-1)
n^5-n = n(n^2+1)(n+1)(n-1)

2. Hi
Originally Posted by 1+1bob
n^4-n = n(n^3-1)
Notice that 1 is a root of $n^3-1$ hence, $n^4-n=n(n-1)(an^2+bn+c)$.
n^5-n = n(n^2+1)(n+1)(n-1)
$n^2+1=(n-\imath)(n+\imath)$ but you may want to factor in $\mathbb{R}$ ? In this case you can't go further (or farther ?) than $n^2+1$.

3. Well, I think I have to use R in this assignment:

"Factorize the expression P(n)=n^x - n for x = 2,3,4,5. Determine if the expression is always divisible by the corresponding x. If divisible use mathematical induction to prove your result by showing whether P(k + 1) - P(k) is always divisible by x. Using appropriate technology, explore more cases and make a conjecture for when n^x - n is divisible by x."

Notice that 1 is a root of hence, .
What do a,b,c do?

cause n(n - 1)(n^2 + n) = n^4 - n^2 ?

4. You should use $c$ to make both $-n^2$ disappear and $n$ appear : developing $n^4-n=n(n-1)(an^2+bn+c)$ and identifying the coefficients yields $a=b=c=1$ hence $n^4-n=n(n-1)(n^2+n+1)$.

5. Originally Posted by 1+1bob
Well, I think I have to use R in this assignment:

"Factorize the expression P(n)=n^x - n for x = 2,3,4,5. Determine if the expression is always divisible by the corresponding x. If divisible use mathematical induction to prove your result by showing whether P(k + 1) - P(k) is always divisible by x. Using appropriate technology, explore more cases and make a conjecture for when n^x - n is divisible by x."

What do a,b,c do?

cause n(n - 1)(n^2 + n) = n^4 - n^2 ?
This is an interesting way to pose the question. The problem setter wants you to discover a wonderful idea on your own. I hope other people will not spoil it and tell you the idea.

"Factorize the expression P(n)=n^x - n for x = 2,3,4,5. Determine if the expression is always divisible by the corresponding x"

1) Does 2 always divide $n^2 - n$?
Hint: Try some values to be sure before we try to guess the result.n^2 - n = n(n-1). Look at the parity of n and n-1.

2) Does 3 always divide $n^3 - n$?
Hint: $n^3 - n = n(n-1)(n+1)$.
Try some values to be sure before we try to guess the result.Consider these questions:What if n was divisible by 3, What if n left a remainder of 1 when divided by 3?, what if n left a remainder of -1?

3)Does 4 always divide $n^4 - n$?
Hint:Lets try some values to be sure before we try to guess the result. if n = 2, 4 does not divide $2^4 - 2$

4) Does 5 always divide $n^5 - n$?
Hint: Try some values to be sure before we try to guess the result.Just observe that n^5 and n will have the same units place. So the difference is divisible by 10 and hence 5.

5) Does 6 always divide $n^6 - n$?
Hint: Lets try some values to be sure before we try to guess the result. if n = 2, 6 does not divide $2^6 - 2$

Hmm can you see a pattern? Now can you establish it using induction?

6. Alright, so I did what you said and found that the stuff will be divisible as long as x is a prime number. I didn't need the factorization for this though, so I assume the factorized form will be used in the proof? Anyways, I'm gonna try to write a proof of this now. Thanks for the help all, I guess I will need more pretty soon though.

7. Originally Posted by 1+1bob
Alright, so I did what you said and found that the stuff will be divisible as long as x is a prime number. I didn't need the factorization for this though, so I assume the factorized form will be used in the proof? Anyways, I'm gonna try to write a proof of this now. Thanks for the help all, I guess I will need more pretty soon though.
You are welcome
So remember you have to prove that $p|(n+1)^p - (n+1)$ using the fact that $p|n^p - n$ for a particular prime p.

Hint: Does $p \bigg{|} {p \choose k}$