The question is already answered (twice). It is however perhaps useful to make a comment about polynomials of degree less than $2$, since the poster seems to believe they are an exception. Such polynomials are certainly uninteresting in this context, but they are not an exception.
Below, we will tacitly assume that we are dealing with polynomial with real coefficients, though with the appropriate algebraic definition of the derivative, one can prove a more general result.
Theorem: Let $P(x)$ be a polynomial. Then the real number $a$ is an (at least) double root of $P(x)$ if and only if $P(a)=P'(a)=0$.
(The "at least" is there because for example $(x-1)^3$ has a triple root at $x=1$. There are two possible interpretations of the term "double root": multiplicity at least $2$ or multiplicity exactly $2$. I am just being careful.)
The theorem holds for all polynomials, without exception. For example, let $P(x)=x-17$. It is true that for any $a$, if $a$ is a double root of $P(x)$, then $P(a)=P'(a)=0$, for $P(x)$ has no double roots. And it is true that if $P(a)=P'(a)=0$, then $a$ is a double root of $P(x)$, since in fact there is no $a$ such that $P(a)=P'(a)=0$.
The same remark would hold for polynomials of degree $0$ (non-zero constants). Finally, let us look at the $0$ polynomial, which in algebra is usually said to have no degree, or degree $-1$, or degree $-\infty$. For this trivial polynomial, every $a$ is a root of multiplicity at least $2$, indeed of infinite multiplicity. And it is true that for every $a$, $P(a)=P'(a)=0$.