Given: \( f(x)=x^{3}+p x^{2}+5 q x+16 \) and - \( (x+2) \) is a factor. When \( (x-3) \) is divided into \( f(x) \), it gives a remainder of -50

To solve the problem, we need to use the information given about the polynomial \( f(x) = x^3 + p x^2 + 5 q x + 16 \). ### Step 1: Use the factor condition Since \( (x + 2) \) is a factor of \( f(x) \), we can use the Factor Theorem, which states that if \( (x - a) \) is a factor of \( f(x) \), then \( f(a) = 0 \). Here, we have: \[ f(-2) = 0 \] Substituting \( x = -2 \) into \( f(x) \): \[ f(-2) = (-2)^3 + p(-2)^2 + 5q(-2) + 16 = 0 \] Calculating this gives: \[ -8 + 4p - 10q + 16 = 0 \] Simplifying: \[ 4p - 10q + 8 = 0 \quad \text{(Equation 1)} \] ### Step 2: Use the remainder condition When \( (x - 3) \) is divided into \( f(x) \), it gives a remainder of -50. Again, using the Remainder Theorem: \[ f(3) = -50 \] Substituting \( x = 3 \) into \( f(x) \): \[ f(3) = (3)^3 + p(3)^2 + 5q(3) + 16 = -50 \] Calculating this gives: \[ 27 + 9p + 15q + 16 = -50 \] Simplifying: \[ 9p + 15q + 43 = -50 \] \[ 9p + 15q = -93 \quad \text{(Equation 2)} \] ### Step 3: Solve the system of equations Now we have a system of two equations: 1. \( 4p - 10q + 8 = 0 \) 2. \( 9p + 15q = -93 \) We can solve these equations simultaneously. Let's express \( p \) from Equation 1: \[ 4p = 10q - 8 \implies p = \frac{10q - 8}{4} = \frac{5q - 4}{2} \] Now, substitute \( p \) into Equation 2: \[ 9\left(\frac{5q - 4}{2}\right) + 15q = -93 \] Multiplying through by 2 to eliminate the fraction: \[ 9(5q - 4) + 30q = -186 \] Expanding: \[ 45q - 36 + 30q = -186 \] Combining like terms: \[ 75q - 36 = -186 \] Adding 36 to both sides: \[ 75q = -150 \] Dividing by 75: \[ q = -2 \] ### Step 4: Substitute back to find \( p \) Now substitute \( q = -2 \) back into the expression for \( p \): \[ p = \frac{5(-2) - 4}{2} = \frac{-10 - 4}{2} = \frac{-14}{2} = -7 \] ### Final Result Thus, the values of \( p \) and \( q \) are: \[ p = -7, \quad q = -2 \]