1. Let's state the problem: We want to find when $3\cdot(x+1)$ is a factor of $x^n + 1$. 2. Note that to check if $(x+1)$ is a factor of $x^n + 1$, by the Factor Theorem, $(x+1)$ is a factor if substituting $x = -1$ yields zero: $$(-1)^n + 1 = 0$$ 3. Simplify this expression: $$(-1)^n + 1 = 0 \implies (-1)^n = -1$$ 4. We know that $(-1)^n = -1$ when $n$ is an odd integer. 5. Therefore, $(x+1)$ is a factor of $x^n + 1$ only when $n$ is odd. 6. The factor $3$ does not affect the divisibility condition regarding $(x+1)$, so the overall product $3\cdot(x+1)$ divides $x^n + 1$ only if $(x+1)$ divides it, i.e., if $n$ is odd. **Final answer:** $3\cdot(x+1)$ is a factor of $x^n + 1$ only if $n$ is an odd integer.