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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 \( q \).

Ask by Evans Hill. in South Africa
Mar 13,2025

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Answer

\( p = 3 \) and \( q = -5 \).

Solution

We are given that the polynomial \[ p x^3 - 8 x^2 + q x + 6 \] is exactly divisible by \[ x^2 - 2x - 3. \] Since the divisor is of degree 2, its roots must also be roots of the cubic polynomial. First, factor the quadratic: \[ x^2 - 2x - 3 = (x - 3)(x + 1). \] Thus, the roots are \( x = 3 \) and \( x = -1 \). ### Step 1: Substitute \( x = 3 \) Substitute into the polynomial: \[ p(3)^3 - 8(3)^2 + q(3) + 6 = 0. \] Calculating each term: \[ 27p - 72 + 3q + 6 = 0. \] Simplify: \[ 27p + 3q - 66 = 0. \] Divide the entire equation by 3: \[ 9p + q - 22 = 0 \quad \Longrightarrow \quad 9p + q = 22. \] ### Step 2: Substitute \( x = -1 \) Substitute \( x = -1 \) into the polynomial: \[ p(-1)^3 - 8(-1)^2 + q(-1) + 6 = 0. \] Calculating each term: \[ - p - 8 - q + 6 = 0. \] Simplify: \[ - p - q - 2 = 0 \quad \Longrightarrow \quad p + q = -2. \] ### Step 3: Solve the System of Equations We now have the system: \[ \begin{aligned} 9p + q &= 22, \\ p + q &= -2. \end{aligned} \] Subtract the second equation from the first: \[ (9p + q) - (p + q) = 22 - (-2), \] \[ 8p = 24 \quad \Longrightarrow \quad p = 3. \] Substitute \( p = 3 \) into \( p + q = -2 \): \[ 3 + q = -2 \quad \Longrightarrow \quad q = -5. \] ### Final Answer \[ \boxed{p = 3, \quad q = -5.} \]

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The Deep Dive

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: \[ \boxed{3} \quad \text{and} \quad \boxed{-5} \]

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