If (x + 1) and (x - 1) are the factors of polynomial `ax^3 + bx^2 + 3x + 5` , then the value of a and b ?

Options

A. a = - 2, b = - 3
B. a= 3, b = 5
C. a= 2, b = 3
D. a = - 3, b = - 5

Answer

To find the values of \( a \) and \( b \) such that \( (x + 1) \) and \( (x - 1) \) are factors of the polynomial \( ax^3 + bx^2 + 3x + 5 \), we can follow these steps:

### Step 1: Use the Factor Theorem
Since \( (x + 1) \) is a factor, substituting \( x = -1 \) into the polynomial should yield 0.

\[ P(-1) = a(-1)^3 + b(-1)^2 + 3(-1) + 5 = 0 \]

### Step 2: Simplify the Equation
Substituting \( x = -1 \):

\[ P(-1) = -a + b - 3 + 5 = 0 \]

This simplifies to:

\[ -a + b + 2 = 0 \]

Rearranging gives us:

\[ b - a + 2 = 0 \quad \text{(Equation 1)} \]

### Step 3: Use the Factor Theorem Again
Now, since \( (x - 1) \) is also a factor, substituting \( x = 1 \) into the polynomial should also yield 0.

\[ P(1) = a(1)^3 + b(1)^2 + 3(1) + 5 = 0 \]

### Step 4: Simplify the Second Equation
Substituting \( x = 1 \):

\[ P(1) = a + b + 3 + 5 = 0 \]

This simplifies to:

\[ a + b + 8 = 0 \]

Rearranging gives us:

\[ a + b + 8 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have a system of equations:

1. \( b - a + 2 = 0 \) (Equation 1)
2. \( a + b + 8 = 0 \) (Equation 2)

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

\[ b = a - 2 \]

### Step 6: Substitute into Equation 2
Substituting \( b = a - 2 \) into Equation 2:

\[ a + (a - 2) + 8 = 0 \]

This simplifies to:

\[ 2a + 6 = 0 \]

### Step 7: Solve for \( a \)
Solving for \( a \):

\[ 2a = -6 \implies a = -3 \]

### Step 8: Find \( b \)
Now substitute \( a = -3 \) back into the expression for \( b \):

\[ b = -3 - 2 = -5 \]

### Final Values
Thus, the values of \( a \) and \( b \) are:

\[ a = -3, \quad b = -5 \]