1. **Problem Statement:** When divided by $x - 3$, the polynomials $x^3 - px^2 + x + 6$ and $2x^3 - x^2 - (p + 3)x - 6$ leave the same remainder. Find the value of $p$. 2. **Formula and Theorem Used:** According to the Remainder Theorem, the remainder when a polynomial $f(x)$ is divided by $x - a$ is $f(a)$. 3. **Step-by-step Solution:** - For the first polynomial $f(x) = x^3 - px^2 + x + 6$, the remainder when divided by $x - 3$ is $f(3)$: $$f(3) = 3^3 - p \times 3^2 + 3 + 6 = 27 - 9p + 3 + 6 = 36 - 9p$$ - For the second polynomial $g(x) = 2x^3 - x^2 - (p + 3)x - 6$, the remainder when divided by $x - 3$ is $g(3)$: $$g(3) = 2 \times 3^3 - 3^2 - (p + 3) \times 3 - 6 = 2 \times 27 - 9 - 3p - 9 - 6 = 54 - 9 - 3p - 9 - 6 = 30 - 3p$$ - Since the remainders are equal: $$36 - 9p = 30 - 3p$$ - Rearranging: $$36 - 30 = 9p - 3p$$ $$6 = 6p$$ $$p = 1$$ 4. **Answer:** The value of $p$ is $1$. This means both polynomials leave the same remainder when divided by $x - 3$ only if $p = 1$.