Question

If the polynomial \( ax^2 + bx + 5 \) leaves a remainder 3 when divided by \( x - 1 \), and a remainder 2 when divided by \( x + 1 \), then \( 2b - 4a \) equals ________

Hint:

The Remainder Theorem is a fundamental tool for problems involving polynomial division. Always remember that dividing by \( (ax-b) \) gives a remainder of \( P(b/a) \). This can save time compared to performing long division.

Answer 1

Correct Answer: 11

Step 1: Understanding the Concept:
This problem uses the Remainder Theorem, which states that if a polynomial P(x) is divided by a linear factor (x - c), the remainder is P(c).
Step 2: Key Formula or Approach:
1. Let \( P(x) = ax^2 + bx + 5 \). 2. Apply the Remainder Theorem for the divisor \( x - 1 \): The remainder is \( P(1) = 3 \). 3. Apply the Remainder Theorem for the divisor \( x + 1 \): The remainder is \( P(-1) = 2 \). 4. Solve the resulting system of linear equations for a and b. 5. Calculate the value of the expression \( 2b - 4a \).
Step 3: Detailed Explanation:
According to the Remainder Theorem, when \( P(x) = ax^2 + bx + 5 \) is divided by \( x - 1 \), the remainder is \( P(1) \). We are given that this remainder is 3. \[ P(1) = a(1)^2 + b(1) + 5 = 3 \] \[ a + b + 5 = 3 \] \[ a + b = -2 \quad \text{(Equation 1)} \] Similarly, when \( P(x) \) is divided by \( x + 1 \) (which is \( x - (-1) \)), the remainder is \( P(-1) \). We are given that this remainder is 2. \[ P(-1) = a(-1)^2 + b(-1) + 5 = 2 \] \[ a - b + 5 = 2 \] \[ a - b = -3 \quad \text{(Equation 2)} \] Now we solve the system of linear equations for a and b. Adding Equation 1 and Equation 2: \[ (a + b) + (a - b) = -2 + (-3) \] \[ 2a = -5 \implies a = -\frac{5}{2} \] Substituting the value of a into Equation 1: \[ -\frac{5}{2} + b = -2 \] \[ b = -2 + \frac{5}{2} = \frac{-4+5}{2} = \frac{1}{2} \] Now we need to find the value of \( 2b - 4a \). \[ 2b - 4a = 2\left(\frac{1}{2}\right) - 4\left(-\frac{5}{2}\right) \] \[ = 1 - (-10) = 1 + 10 = 11 \] Step 4: Final Answer:
The value of \( 2b - 4a \) is 11.