The polynomial $x^5-a^2{x}^3+2x+a+3$ has $(x-a)$ among its factors. Then a equals:

By the factor theorem the value at $x=a$ must be zero: $a^5-a^2\cdot a^3+2a+a+3=a^5-a^5+3a+3=3a+3$. Setting $3a+3=0$ gives $a=-1$. (Solving $3a=3$ instead of $3a=-3$ gives 1; overlooking the $2a$ contributed by the term $2x$ leaves $a+3=0$ and the wrong value −3, and 3 flips that sign.) Check with $a=-1:x^5-x^3+2x+2$ at $x=-1$ is −1 + 1 − 2 + 2 = 0.