Factorise : `n^(3)-3n - 2 ,` given that `n+1` is a factor .The factors are

Options

A. `(n+1)^(2)(n-2)`
B. `(n+1)(n-2)^(2)`
C. `(n-1)^(2)(n+2)`
D. None of these

Answer

To factorise the expression \( n^3 - 3n - 2 \) given that \( n + 1 \) is a factor, we will follow these steps:

### Step 1: Polynomial Division
Since \( n + 1 \) is a factor, we will divide \( n^3 - 3n - 2 \) by \( n + 1 \).

1. Set up the division:
\[ \text{Divide } n^3 - 3n - 2 \text{ by } n + 1. \]

### Step 2: Perform the Division
1. Divide the leading term \( n^3 \) by \( n \):
\[ n^3 \div n = n^2. \]
2. Multiply \( n^2 \) by \( n + 1 \):
\[ n^2(n + 1) = n^3 + n^2. \]
3. Subtract this from the original polynomial:
\[ (n^3 - 3n - 2) - (n^3 + n^2) = -n^2 - 3n - 2. \]

### Step 3: Continue the Division
1. Now, divide the leading term \(-n^2\) by \(n\):
\[ -n^2 \div n = -n. \]
2. Multiply \(-n\) by \(n + 1\):
\[ -n(n + 1) = -n^2 - n. \]
3. Subtract this from the previous result:
\[ (-n^2 - 3n - 2) - (-n^2 - n) = -2n - 2. \]

### Step 4: Final Division Step
1. Divide \(-2n\) by \(n\):
\[ -2n \div n = -2. \]
2. Multiply \(-2\) by \(n + 1\):
\[ -2(n + 1) = -2n - 2. \]
3. Subtract this from the previous result:
\[ (-2n - 2) - (-2n - 2) = 0. \]

### Step 5: Write the Result
The division gives us:
\[ n^3 - 3n - 2 = (n + 1)(n^2 - n - 2). \]

### Step 6: Factor the Quadratic
Now we need to factor \( n^2 - n - 2 \):
1. Look for two numbers that multiply to \(-2\) and add to \(-1\). These numbers are \(-2\) and \(1\).
2. Thus, we can factor it as:
\[ n^2 - n - 2 = (n - 2)(n + 1). \]

### Final Step: Combine the Factors
Putting it all together, we have:
\[ n^3 - 3n - 2 = (n + 1)(n - 2)(n + 1) = (n + 1)^2(n - 2). \]

### Conclusion
The factors of \( n^3 - 3n - 2 \) are:
\[ (n + 1)^2(n - 2). \]