Definition of Associates for commutative rings with identity: Let a, b $\in$ R. Then b is an associate of a if b = au for some unit u of R.
With this definition, I am able to conclude that the associates of $(x+1)$ in $\mathbb{Q}$[x] are $(x+1)$ $\cdot$ u where u $\in$ R. Thus, is it sufficient to say that $(x^2 + x)$ is an associate of $(x+1)$ in $\mathbb{Q}$[x] since $(x^2 + x) = x(x + 1)$?
A unit in $\mathbb Q[x]$ is an invertible element of $\mathbb Q[x]$ i.e. $f \in \mathbb Q[x]$ is a unit if there exists $g \in \mathbb Q[x]$ such that $fg = 1$, where $1$ is the constant polynomial with value $1$.
From here, it is easy to see that the only units in $\mathbb Q[x]$ are the non-zero constant polynomials : because if $\deg f > 0$ then $\deg fg \geq \deg f > 0$ so $fg \neq 1$. Therefore, $f$ is a constant polynomial, and obviously then $f = c \implies g = \frac 1c$ works whenever $c \neq 0$.
Thus, the associates of $x+1$ are of the form $c(x+1)$ for any $c \in \mathbb Q, c \neq 0$. Examples would be $2(x+1)$, or $\frac 12(x+1)$ or $-\pi(x+1)$.