Need help solving n^5+80=5n^4+16n?

[ymersion]

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

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

 

[CompuChip]

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

 

[Borek]

Does n mean you are looking only for integer roots?

 

[ymersion]

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

 

[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.

 

[ymersion]

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

 

[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.

 

[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 that's one of the roots then you can continue to find the other by doing your long division.

 

[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

 

[Borek]

http://en.wikipedia.org/wiki/Rational_root_theorem

 

[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.

 

[Borek]

There are many ways to skin that cat