Let \( P(x) \) be a quadratic polynomial such that: \( P(0) = 1 \). \( P(x) \) leaves a remainder of 4 when divided by \( x - 1 \). \( P(x) \) leaves a remainder of 6 when divided by \( x + 1 \). Find the polynomial \( P(x) \).
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, \( {P(-2) = 19} \)
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