The expression \( p x^{3}-8 x^{2}+q x+6 \) is exactly divisible by \( x^{2}-2 x-3 \) Find the values of \( p \) and \( q \).

To determine the values of \( p \) and \( q \) such that the expression \( p x^{3}-8 x^{2}+q x+6 \) is exactly divisible by \( x^{2}-2 x-3 \), we start by factoring \( x^{2}-2 x-3 \). The quadratic can be factored as: \[ x^{2}-2 x-3 = (x-3)(x+1) \] For the polynomial \( p x^{3}-8 x^{2}+q x+6 \) to be divisible by \( x^{2}-2 x-3 \), it must be zero at both roots \( x = 3 \) and \( x = -1 \). We substitute these values into the polynomial and set the equations to zero. 1. For \( x = 3 \): \[ p(3)^{3} - 8(3)^{2} + q(3) + 6 = 0 \] \[ 27p - 72 + 3q + 6 = 0 \] \[ 27p + 3q - 66 = 0 \] This simplifies to: \[ 9p + q = 22 \quad \text{(Equation 1)} \] 2. For \( x = -1 \): \[ p(-1)^{3} - 8(-1)^{2} + q(-1) + 6 = 0 \] \[ -p - 8 - q + 6 = 0 \] \[ -p - q - 2 = 0 \] This simplifies to: \[ p + q = -2 \quad \text{(Equation 2)} \] Now, we can solve the system of equations represented by Equation 1 and Equation 2. From Equation 2, we have: \[ q = -2 - p \] Substituting this expression for \( q \) into Equation 1: \[ 9p + (-2 - p) = 22 \] \[ 9p - p - 2 = 22 \] \[ 8p - 2 = 22 \] \[ 8p = 24 \] \[ p = 3 \] Now substituting \( p = 3 \) back into Equation 2: \[ 3 + q = -2 \] \[ q = -2 - 3 = -5 \] Thus, the values of \( p \) and \( q \) are: \[ \boxed{3} \quad \text{and} \quad \boxed{-5} \]