Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$


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*Find the quotient of $P(x)$ divided by $x^2 + x + 1$

*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers. 
I've done the first task and came up with $P(x)= (x+1)^5 + x = (x^3+4x^2+5x+1)(x^2+x+1)$ which can be written as $A =BC +D$ where $D=0$. So, for the second task I think I have to prove that B or C or both are prime, but I don't know how to continue.

$B$ and $C$ aren't necessarily prime, but they have distinct factors.  
To show this, continue as follows:
$(x^3+4x^2+5x+1)=(x^2+x+1)(x+3)+x-2.$
$x^2+x+1=(x-2)(x+3)+7.$
Therefore, $\gcd(x^3+4x^2+5x+1,x^2+x+1)|7,$ so there are two possibilities.
Case ($1$):  $\gcd(x^3+4x^2+5x+1,x^2+x+1)=1.$
In this case, since, for positive $x,$ $x^3+4x^2+5x+1>x^2+x+1>1$, 
$x^3+4x^2+5x+1$ and $x^2+x+1$ have distinct prime factors.
Case ($2$):  $\gcd(x^3+4x^2+5x+1,x^2+x+1)=7 .$
In this case $7|x-2.$  For $x=2, x^3+4x^2+5x+1=35$ has two distinct prime factors ($5$ and $7$).  
For $x>2$, $x^3+4x^2+5x+1>x^2+x+1>7$, and $\gcd\left(\dfrac {x^3+4x^2+5x+1}7,\dfrac{x^2+x+1}7\right)=1,$
so $\dfrac {x^3+4x^2+5x+1}7$  and $\dfrac{x^2+x+1}7$ have distinct prime factors.
