Proof of existing degree $n$ binomial 
Let $P(x)$ be a polynomial with real coefficients such that $P(x) > 0$ for all $x \ge 0$. Prove that there exists a positive integer $n$ such that $(x + 1)^n P(x)$ is a polynomial with nonnegative coefficients.

HINTS ONLY.
I tried the binomial theorem with a quadratic case with:
$f(x) = (x+1)^n (x^2 + 2x +1) = \sum_{k=0}^{n} \binom{n}{k} (x^{k+2} - x^{k+1} + 80x^k)$
But nothing really past this.
 A: Let ${\cal S}$ be the set of  polynomials $P$ such that there exists a positive integer $n$ such that the coefficients of $(1+X)^nP$ are all positive. 


*

*${\cal S}$ is closed under polynomial multiplication, since  the set of polynomials with positive coefficients is closed under polynomial multiplication.

*${\cal S}$ contains constant positive polynomials.

*${\cal S}$ contains all polynomials of the form $X+\lambda$ with $\lambda>0$.

*${\cal S}$ contains all polynomials of the form $X^2-2\alpha X+\beta$ with $\alpha^2<\beta$. Indeed, there is nothing to be proved 
if $\alpha\leq0$ so we may suppose that $\alpha>0$. Now, let us consider
$$
(X+1)^n(X^2-2\alpha X+\beta)=X^{n+2}+\sum_{k=0}^{n}c_kX^{k+1}+\beta
$$
with 
$$ c_k=\binom{n}{k-1}-2\alpha \binom{n}{k}+\beta\binom{n}{k+1}
=\binom{n}{k}\left(\frac{k}{n-k+1}+\beta\frac{n-k}{k+1}-2\alpha
\right)$$
that is $c_k=\binom{n}{k}F\left(\frac{k}{n}\right)$ where
$$F(x)=\frac{x}{1+\frac{1}{n}-x}+\beta \frac{1-x}{\frac{1}{n}+x}-2\alpha,\quad\hbox{for $0\leq x\leq 1$.} $$
Now, suppose that $n>\max\left(\sqrt{\beta},\sqrt{1/\beta}\right) $, so that $F$ attains its minimum on the interval $[0,1]$ at
$$x_0=\frac{\sqrt{\beta}(1+n)-1}{ (1+\sqrt{\beta})n}$$ and the minimum of $F$ on $[0,1]$ is
$$F(x_0)=2(\sqrt{\beta}-\alpha)-\frac{(1+\sqrt{\beta})^2}{n+2}$$
It follows that if
$$n> n_0= \max\left(\frac{(1+\sqrt{\beta})^2}{2(\sqrt{\beta}-\alpha)},\sqrt{\beta },\frac{1}{\sqrt{\beta}} \right)$$
then $F(x)$ is positive for every $x\in[0,1]$. That is $c_k>0$ for $0\leq k\leq n$. 
So, we have shown that $(1+X)^n(X^2-2\alpha X+\beta)$ has 
positive coefficients if $n$ is large enough and the proof of this point is complete.

*${\cal S}$ contains all polynomials having no nonnegative zeros. Indeed, such a polynomial is 
the product of polynomials each one of them has one of the  forms  discussed earlier. The desired conclusion follows.

