1. **Problem Statement:** We know that when a polynomial $f(x)$ is divided by $x - 1$, the remainder is $R$. We want to find the remainder when $f(x + 1)$ is divided by one of the given options. 2. **Given Information:** - $f(x) = (x - 1)Q(x) + R$ where $Q(x)$ is the quotient. - By the Remainder Theorem, $f(1) = R$. 3. **Step to find remainder when dividing $f(x + 1)$:** We want to find the remainder when $f(x + 1)$ is divided by a linear polynomial. The remainder when dividing by a linear polynomial $x - a$ is $f(a)$. 4. **Check each divisor:** - For $x - 1$: remainder is $f(1 + 1) = f(2)$. - For $x + 1$: remainder is $f(-1 + 1) = f(0)$. - For $x + 2$: remainder is $f(-2 + 1) = f(-1)$. 5. **Given that the remainder is also $R$ when dividing $f(x + 1)$, it means:** $$f(\text{value}) = R$$ 6. **From the note $x + 1 = 1 \Rightarrow x = 0$, so dividing $f(x + 1)$ by $x - 1$ corresponds to evaluating $f(2)$, which is not necessarily $R$. But dividing $f(x + 1)$ by $x$ (which is $x - 0$) corresponds to evaluating $f(0)$. 7. **Since the remainder is $R$ when dividing $f(x + 1)$ by $x - 0$ (or $x$), the divisor must be $x - 0$, which is $x$ or equivalently $x + 0$. 8. **Among the options, $x + 1$ corresponds to $x - (-1)$, so dividing by $x + 1$ means evaluating $f(-1 + 1) = f(0)$, which matches the condition. **Final answer:** The remainder is $R$ when $f(x + 1)$ is divided by $x + 1$. $$\boxed{x + 1}$$