Solve n^5+80=5n^4+16n

1. Sep 9, 2008

ymersion

Can someone help me solve this? I can't figure it out.

n^5+80=5n^4+16n

2. Sep 9, 2008

CompuChip

Factorisation might work here. Especially if you can "guess" one root.

3. Sep 9, 2008

Staff: Mentor

Does n mean you are looking only for integer roots?

4. Sep 9, 2008

ymersion

n is just the variable, any rational answer will work. I've tried to factor but, I can't.

5. Sep 9, 2008

CompuChip

Maybe if you can guess a root (try n = 1, 2, 3, -1, -2, -3, ...), you can factor the remaining polynomial.
Especially for homework assignments this tends to work, because the exercise is constructed such that most roots are integers near 0.

6. Sep 9, 2008

ymersion

I have the answer. I just don't know how to solve it. Answer is n=5,2,-2

7. Sep 9, 2008

snipez90

Move everything to one side and note that 80 = 5*16. Factor into a sum of two terms each involving a product of two terms. Then try to factor the resulting expression.

Last edited: Sep 9, 2008
8. Sep 9, 2008

virtueboy15

Hey,you have to guess a value of n(which i found to be -2) to make n^5-80-5n^4+16n equal to zero and thats one of the roots then you can continue to find the other by doing your long division.

9. Sep 9, 2008

ymersion

I finally got it. Thanks for the help. FYI, this is how I solved it.

n^5-5n^4-16n+80=0
n^4(n-5)-16(n-5)=0
(n^4-16)(n-5)=0
n^4=16 and n-5=0
n=+/-2,5

10. Sep 9, 2008

Staff: Mentor

11. Sep 9, 2008

snipez90

That was my first hunch too. Then I realized with the constant term being 80, factoring was the best way to go. Keeping in mind Compuchip's hint, it's easy to see -1, 0, and 1 won't work. 2 worked out nicely and really at that point synthetic division would have done the trick but I remembered learning "factoring by grouping" and realizing that 80 was a multiple of 16 made everything clear.

12. Sep 9, 2008

Staff: Mentor

There are many ways to skin that cat