Step-by-step Solution:

We are given:
\( P(0) = 1 \)
The remainder when \( P(x) \) is divided by \( x - 1 \) is 4 → \( P(1) = 4 \)
The remainder when \( P(x) \) is divided by \( x + 1 \) is 6 → \( P(-1) = 6 \)
Let the quadratic polynomial be: \[ P(x) = ax^2 + bx + c \] Step 1: Use the given values
From \( P(0) = 1 \), we get: \[ P(0) = a(0)^2 + b(0) + c = c = 1 \] So, \( c = 1 \) Now use \( P(1) = 4 \): \[ a(1)^2 + b(1) + 1 = 4 \Rightarrow a + b + 1 = 4 \Rightarrow a + b = 3 \quad \text{(1)} \] Now use \( P(-1) = 6 \): \[ a(-1)^2 + b(-1) + 1 = 6 \Rightarrow a - b + 1 = 6 \Rightarrow a - b = 5 \quad \text{(2)} \] Step 2: Solve the system of equations
Add equations (1) and (2): \[ (a + b) + (a - b) = 3 + 5 \Rightarrow 2a = 8 \Rightarrow a = 4 \] Substitute \( a = 4 \) into (1): \[ 4 + b = 3 \Rightarrow b = -1 \] So, the polynomial is: \[ P(x) = 4x^2 - x + 1 \] Now, plug in \( x = -2 \): \[ P(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 2 + 1 = {19} \] ✅ So, \( {P(-2) = 19} \)