Show that the polynomial $P(x)=(x^4+x^3-3x^2+x+2)^n$ has negative coefficients 
Show that for any positive integer $n$, the polynomial $$P(x)=(x^4+x^3-3x^2+x+2)^n$$ has negative coefficients.

For $n=1$ it is clear and for $n=2$ it's true because $$p(x)=x^8+2x^7-5x^6-4x^5+\cdots$$
which also have negative coefficients. But for any positive integer $n$, I can't prove it.
 A: Note that the sum of the coefficients of a polynomial $Q(x)$ is just $Q(1)$. So the sum of all of the coefficients of $P(x)^n$ is
$$P(1)^n=2^n$$
Now, consider the coefficients of $x^{4n}$ and $x^0$ in $P(x)^n$. Note that the first term can only be made, when expanding the product of $n$ polynomials, by multiplying all of the $x^4$ terms, and the second term can only be made by multiplying all of the $x^0$ terms. 
Thus, the coefficient of $x^{4n}$ in $P(x)^n$ is $1$, and the coefficient of $x^0$ in $P(x)$ is $2^n$. 
However, these sum to $2^n+1>2^n$, so the other coefficents must sum to $-1$. Because of this, at least one must be negative.
A: The polynomial $p(x)=x^4+x^3-3x^2+x+2$ has the property $p(0)=p(1)$, hence by Rolle's theorem there is a zero of $p'(x)$ in the interval $(0,1)$. The same argument applies to $P(x)=p(x)^n$ where $P(0)=2^n$ and $P'(0)=n2^{n-1}$: if the coefficients of $P(x)$ were all non-negative, it would be impossible for $P'(x)$ to have a zero in $(0,1)$, contradiction.
