Example Question - constant values

Solving for Constants in a Polynomial Function

The image contains a question that asks you to find the values of constants 'a' and 'b' in a polynomial function \( P(x) = 3x^3 + ax^2 + bx - 6 \), given that \( x - 2 \) is a factor of \( P(x) \) and that the remainder is 27 when \( P(x) \) is divided by \( (x + 2) \). To solve this: 1. If \( x - 2 \) is a factor of \( P(x) \), it means that \( P(2) = 0 \). 2. If the remainder is 27 when \( P(x) \) is divided by \( x + 2 \), it means that \( P(-2) = 27 \). Let's use these conditions to form two equations to solve for 'a' and 'b'. 1. \( P(2) = 0 \) \[ 3(2)^3 + a(2)^2 + b(2) - 6 = 0 \] \[ 3(8) + 4a + 2b - 6 = 0 \] \[ 24 + 4a + 2b = 6 \] \[ 4a + 2b = 6 - 24 \] \[ 4a + 2b = -18 \] \[ 2a + b = -9 \] (Equation 1) 2. \( P(-2) = 27 \) \[ 3(-2)^3 + a(-2)^2 + b(-2) - 6 = 27 \] \[ 3(-8) + 4a - 2b - 6 = 27 \] \[ -24 + 4a - 2b = 27 + 6 \] \[ 4a - 2b = 33 + 24 \] \[ 4a - 2b = 57 \] \[ 2a - b = 57/2 \] (Equation 2) Now we have a system of linear equations: \[ 2a + b = -9 \] (Equation 1) \[ 2a - b = 57/2 \] (Equation 2) Multiply Equation 1 by 2 and add it to Equation 2: \[ 4a + 2b = -18 \] (Equation 1 transformed) \[ 2a - b = 57/2 \] (Equation 2) \[ (4a + 2b) + (2a - b) = -18 + 57/2 \] \[ 6a + b = 21/2 \] (Equation 3) Now subtract Equation 1 from Equation 3: \[ 6a + b - (2a + b) = 21/2 + 9 \] \[ 4a = 21/2 + 18/2 \] \[ 4a = 39/2 \] \[ a = \frac{39/2}{4} \] \[ a = 39/8 \] \[ a = 4.875 \] Now we can substitute the value of 'a' back into Equation 1 to solve for 'b': \[ 2a + b = -9 \] \[ 2(4.875) + b = -9 \] \[ 9.75 + b = -9 \] \[ b = -9 - 9.75 \] \[ b = -18.75 \] The values are \( a \approx 4.875 \) (or \( a = 39/8 \)) and \( b \approx -18.75 \) (or \( b = -75/4 \)).