Given the polynomial \( P(x) = 2x^3 - 3x^2 + 4x - 5 \) and \( Q(x) = x - 1 \), find the remainder when \( P(x) \) is divided by \( Q(x) \). Additionally, if \( R(x) \) is the remainder and \( \theta \) is the root of \( Q(x) \), find the value of \( \cos(\theta) \) if \( R(\theta) = \cos(\theta) \)
Answer
Step 4: Since it's given that \( R(\theta) = \cos(\theta) \), we have \( \cos(\theta) = -2 \)
Step 1 ：Step 1: Divide \( P(x) \) by \( Q(x) \) using polynomial long division. This gives us \( P(x) = Q(x) \cdot (2x^2 - x + 5) + R(x) \), where \( R(x) \) is the remainder.
Step 2 ：Step 2: Set \( Q(x) = 0 \) to find the root \( \theta \), we have \( x - 1 = 0 \) which gives \( \theta = 1 \)
Step 3 ：Step 3: Substitute \( \theta = 1 \) into \( R(x) \), we get \( R(1) = 2 \cdot 1^3 - 3 \cdot 1^2 + 4 \cdot 1 - 5 = -2 \)
Step 4 ：Step 4: Since it's given that \( R(\theta) = \cos(\theta) \), we have \( \cos(\theta) = -2 \)
