For what values of a and b, `x^4 + ax^3 + 2x^2 - 3x + b` is divisible by `x^2 -1` ?

Options

A. `a=3 , b=4`
B. a=2 , b=4
C. a=3 , b=-3
D. a=1 , b=-3

Answer

To determine the values of \( a \) and \( b \) such that the polynomial \( P(x) = x^4 + ax^3 + 2x^2 - 3x + b \) is divisible by \( x^2 - 1 \), we can use the Remainder Theorem. Since \( x^2 - 1 = (x - 1)(x + 1) \), the polynomial \( P(x) \) must equal zero when \( x = 1 \) and \( x = -1 \).

**Step 1: Evaluate \( P(1) \)**

Substituting \( x = 1 \) into \( P(x) \):
\[ P(1) = 1^4 + a(1^3) + 2(1^2) - 3(1) + b \]
\[ = 1 + a + 2 - 3 + b \]
\[ = a + b + 0 \]
Setting \( P(1) = 0 \):
\[ a + b = 0 \quad \text{(Equation 1)} \]

**Step 2: Evaluate \( P(-1) \)**

Substituting \( x = -1 \) into \( P(x) \):
\[ P(-1) = (-1)^4 + a(-1)^3 + 2(-1)^2 - 3(-1) + b \]
\[ = 1 - a + 2 + 3 + b \]
\[ = -a + b + 6 \]
Setting \( P(-1) = 0 \):
\[ -a + b + 6 = 0 \quad \text{(Equation 2)} \]

**Step 3: Solve the system of equations**

We have the following two equations:
1. \( a + b = 0 \)
2. \( -a + b + 6 = 0 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = -a \]

Substituting \( b = -a \) into Equation 2:
\[ -a + (-a) + 6 = 0 \]
\[ -2a + 6 = 0 \]
\[ -2a = -6 \]
\[ a = 3 \]

Now substituting \( a = 3 \) back into Equation 1 to find \( b \):
\[ 3 + b = 0 \]
\[ b = -3 \]

**Final Values:**
Thus, the values of \( a \) and \( b \) are:
\[ a = 3, \quad b = -3 \]

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