The only constant polynomial that works is $P(x)\equiv 0$, it is easy to see that this works, so from now and on we assume $P$ is non-constant. Claim: $P$ has even degree. Proof: Assume not, then the polynomial $Q(x)=(x^{20}+x^{19})P(x)-P(2x+\alpha)$ has odd degree and hence takes both positive and negative values, wich contradicts the fact that $Q(x)$ is non-negative for all reals. $\blacksquare$ Note that $P$ has a positive leading coefficient. We now split into cases if $\alpha>0$, then we take $x=0$ we get that $P(\alpha)\leq 0$, then we take $x=\alpha$ to get $P(3\alpha)\leq 0$, so for a $k>0$ such that $P(k)\leq 0$ we take $x=k$ to get $P(2k+\alpha) \leq 0$, so this means that there exists a $t$ such that for all $x\geq t$ we have $P(x) <0$, because if such $t$ wouldn't exist then $P$ would be identically $0$, but this means that $P$ has negative leading coefficient since it has even degree, which is not true. If $\alpha \leq -1$, then we take $x=0$ to get $P(\alpha)\leq 0$ and $x=\alpha$ to get $P(3\alpha)\leq0$, so like in the above case there exists a negative $t$ such that for all $x \leq t$ we have $P(x)<0$, and with the same argument as above we have no solutions in this case. If $-1<\alpha \leq 0$, we take $x=-1$ to get $P(\alpha-2)\leq 0$ and $x=\alpha-2$ to get $P(3\alpha-4)\leq 0$, and just like in the above cases there exists a $t$ such that for all $x<t$ we have $P(x)\leq 0$ wich again gives us no solutions. After considering all cases we can conclude that $P \equiv 0$ is the only solution. $\blacksquare$

A slightly different solution with the same idea:

Different solution The answer is only zero polynomial work. It's easy to see that only constant polynomial which works is zero. For now, assume that $P$ is non-constant. Consider the polynomial $Q(x) = (x^{20}+x^{19})P(x) - P(2x+\alpha)$. As $Q$ always take positive values, $Q$ must have even degree and positive leading coefficient. This implies $\deg P$ is even and $P$ has positive leading coefficient. Note that $P(\alpha) < 0$ while $P(\infty) = \infty$. Thus by IVT $P$ has at least one root. Let $r$ be the least root and $s$ be the greatest root. Consider the following. By the inequality, $P(2r+\alpha)\leq 0$. Thus by IVT, $P$ has at least one root in $(-\infty,2r+\alpha)$. This means $2r+\alpha\geq r\implies r\geq-\alpha$. Similarly, $P(2s+\alpha)\leq 0$ implies $P$ has at least one root in $(2s+\alpha,\infty)$. This means $2s+\alpha\leq s\implies s\leq -\alpha$. As $r\leq s$, this forces $r=s=-\alpha$. Thus $P$ has exactly one real root which means $P(x)\geq 0$ for all $x$ with equality at $x=-\alpha$. However, plugging in $x=0,-1$ into the inequality gives $P(\alpha) \leq 0$ and $P(\alpha-2) = 0$, which contradicts the equality case proven above.

Answer:$P(x)=0$

The only constant polynomial that works is $P(x)\equiv 0$, so from now we assume $P$ is non-constant. Denote $Q(x)=(x^{20}+x^{19})P(x)-P(2x+\alpha) \geq 0$. Observe that $P$ has even degree, otherwise $Q$ has odd degree and hence attains negative values, contradiction. Similarly, note that $P$ has a positive leading coefficient. Setting $x=0, -1$ yield $P(\alpha) \leq 0$ and $P(\alpha - 2) \leq 0$. Since $P\not\equiv 0$, we have that $P$ has finitely many roots. By the Intermediate Value theorem, at least one of them is in $(-\infty, \alpha - 2]$ and at least one of them is in $[\alpha, \infty)$. Suppose $\alpha > 0$. Let $r \geq \alpha > 0$ be the largest root of $P$. Setting $x=r$ yields $P(2r+\alpha) \leq 0$, so by the Intermediate Value theorem $P$ has a root in $[2r + \alpha, \infty)$. But $2r + \alpha > r$, contradicting the maximality of $r$. Now suppose $\alpha \leq 0$. Let $s \leq \alpha - 2 < 0$ be the smallest root of $P$. Setting $x=s$ yields $P(2s+\alpha) \leq 0$, so by the Intermediate Value theorem $P$ has a root in $(-\infty, 2s+\alpha]$. But $2s + \alpha < s$, as $s+\alpha \leq 0$ for $s<0$, $\alpha < 0$, contradicting the minimality of $s$.

The answer is $P\equiv 0$. Let $Q(x)=(x^{20}+x^{19})P(x)-P(2x+\alpha)$, then $Q(x)\geq 0,\forall x\in \mathbb{R}$. From this, we easily to see that $\deg P$ is even and coefficient of highest degree of $P$ is positive. If $P(x)>0,\forall x\in \mathbb{R}$, then $x^{20}+x^{19}>0,\forall x\in \mathbb{R}$, this is wrong with $x\in (-1;0)$. If $P$ has real root. Since $\deg P$ is even, $P$ will have at least two real roots. Let $r$ is the smallest root of $P$. Then $P(2r+\alpha)\leq 0$, notice that $\deg P$ is even and the coefficient of highest degree of $P$ is positive, so $\lim_{x\to -\infty}P(x)=+\infty$, so there exists a real number $r'$ such that $P(r')=0$ and $r'\in (-\infty;2r+\alpha)$. Notice that $r'<r$, then $2r+\alpha\geq r$, or $a\geq -r$. Let $s$ is the largest root of $P$. Then $P(2s+\alpha)\leq 0$, notice that $\lim_{x\to+\infty}P(x)=+\infty$, so there exists a real number $s'$ such that $P(s')=0$ and $s'\in (2s+\alpha;+\infty)$. So $2s+\alpha\leq s$, or $a\leq -s$. Therefore, $-s\geq a\geq -r$, or $s\geq r$, but from definitions of $r$ and $s$, we have $P$ has only one real root and get this root is even multiple root, so $P(x)\geq 0,\forall x\in \mathbb{R}$. Then $(x^{20}+x^{19})P(x)\geq 0,\forall x\in \mathbb{R}$. In this inequality, for all $x\in (-1;0)$, $P(x)\leq 0$. Then $P(x)\equiv 0$