5. When the expression \( x^{3}+p x^{2}+2 x+1 \) is divided by \( (x-2) \), the remainder is thrice the remainder when the divisor is \( (x+1) \). Find the value of \( p \). [Answer Key] \( p=-19 \)

To solve the problem, we will use the Remainder Theorem, which states that the remainder of the division of a polynomial \( f(x) \) by a linear divisor \( (x - a) \) is \( f(a) \). First, we calculate the remainders when the polynomial \( f(x) = x^3 + px^2 + 2x + 1 \) is evaluated at \( x = 2 \) and \( x = -1 \): 1. **Calculating \( f(2) \)**: \[ f(2) = 2^3 + p(2^2) + 2(2) + 1 = 8 + 4p + 4 + 1 = 4p + 13 \] 2. **Calculating \( f(-1) \)**: \[ f(-1) = (-1)^3 + p(-1)^2 + 2(-1) + 1 = -1 + p - 2 + 1 = p - 2 \] According to the problem, the remainder when dividing by \( (x-2) \) is three times the remainder when dividing by \( (x+1) \). Therefore, we have the equation: \[ 4p + 13 = 3(p - 2) \] Expanding the right side: \[ 4p + 13 = 3p - 6 \] Now, we rearrange the equation to isolate \( p \): \[ 4p - 3p = -6 - 13 \implies p = -19 \] Thus, the value of \( p \) is \( -19 \).