If `(x+1) and (x-1)` are factor of `ax^(3) + bx^(2) + 3x+5`. Find the value of a and b

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that \( (x + 1) \) and \( (x - 1) \) are factors of the polynomial \( ax^3 + bx^2 + 3x + 5 \).

### Step 1: Set up the polynomial function
Let \( f(x) = ax^3 + bx^2 + 3x + 5 \).

### Step 2: Use the factor theorem
Since \( (x + 1) \) is a factor, we substitute \( x = -1 \) into the polynomial and set it equal to zero:
\[ f(-1) = a(-1)^3 + b(-1)^2 + 3(-1) + 5 = 0 \]
This simplifies to:
\[ -a + b - 3 + 5 = 0 \]
\[ -a + b + 2 = 0 \]
Rearranging gives us the first equation:
\[ a - b = 2 \quad \text{(Equation 1)} \]

### Step 3: Substitute \( x = 1 \) for the second factor
Next, since \( (x - 1) \) is also a factor, we substitute \( x = 1 \) into the polynomial and set it equal to zero:
\[ f(1) = a(1)^3 + b(1)^2 + 3(1) + 5 = 0 \]
This simplifies to:
\[ a + b + 3 + 5 = 0 \]
\[ a + b + 8 = 0 \]
Rearranging gives us the second equation:
\[ a + b = -8 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have a system of two equations:
1. \( a - b = 2 \)
2. \( a + b = -8 \)

We can add these two equations together:
\[ (a - b) + (a + b) = 2 - 8 \]
This simplifies to:
\[ 2a = -6 \]
Dividing both sides by 2 gives:
\[ a = -3 \]

### Step 5: Substitute back to find \( b \)
Now substitute \( a = -3 \) back into Equation 2:
\[ -3 + b = -8 \]
Solving for \( b \) gives:
\[ b = -8 + 3 = -5 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ a = -3, \quad b = -5 \]