+0  
 
+1
100
5
avatar+853 

Find the nonconstant polynomial \(P(x)\) such that

\(P(P(x)) = (x^2 + x + 1) P(x)\)
 

 Dec 28, 2018
 #1
avatar+99117 
0

I do not know but it looks interesting. I will be interested in the answer when it comes  laugh

 Dec 29, 2018
 #2
avatar+4394 
+2

it's pretty clear P(x) can't be an odd degree so we'll try a 2nd degree polynomial

 

\(\text{Letting }P(x) = c_0 + c_1 x + c_2 x^2\\ \text{then grinding out the expansions of each side and equating like power of x coefficients}\\ \text{we get two solutions}\\ P(x) = 0\\ P(x) = x+x^2\)

.
 Dec 29, 2018
edited by Rom  Dec 29, 2018
 #3
avatar+99117 
0

Thanks Rom,

I will have to think about that one :)

Melody  Dec 29, 2018
 #4
avatar+4394 
+3

Maybe you can follow this.

 

Rom  Dec 29, 2018
 #5
avatar+99117 
0

Thanks Rom. 

I will look :)

Melody  Dec 29, 2018

9 Online Users

avatar