Using the Remainder theorem, factorise the expression `2x^(3)+x^(2)-2x-1` completely.

Answer

To factorize the expression \(2x^3 + x^2 - 2x - 1\) completely using the Remainder Theorem, we will follow the steps outlined below:

### Step 1: Define the Polynomial
Let \(f(x) = 2x^3 + x^2 - 2x - 1\).

### Step 2: Identify Possible Rational Roots
According to the Remainder Theorem, we can check for possible rational roots by considering the factors of the constant term. The constant term here is \(-1\), and its factors are \(\pm 1\).

### Step 3: Test Possible Roots
We will test \(x = 1\) and \(x = -1\).

1. **Testing \(x = 1\)**:
\[ f(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0 \]
Since \(f(1) = 0\), \(x - 1\) is a factor of the polynomial.

### Step 4: Perform Polynomial Division
Now, we will divide \(f(x)\) by \(x - 1\) using synthetic division or long division.

1. **Long Division**:
- Divide \(2x^3\) by \(x\) to get \(2x^2\).
- Multiply \(2x^2\) by \(x - 1\) to get \(2x^3 - 2x^2\).
- Subtract:
\[ (2x^3 + x^2) - (2x^3 - 2x^2) = 3x^2 \]
- Bring down the next term \(-2x\) to get \(3x^2 - 2x\).
- Divide \(3x^2\) by \(x\) to get \(3x\).
- Multiply \(3x\) by \(x - 1\) to get \(3x^2 - 3x\).
- Subtract:
\[ (3x^2 - 2x) - (3x^2 - 3x) = x \]
- Bring down \(-1\) to get \(x - 1\).
- Divide \(x\) by \(x\) to get \(1\).
- Multiply \(1\) by \(x - 1\) to get \(x - 1\).
- Subtract:
\[ (x - 1) - (x - 1) = 0 \]

Thus, we have:
\[ f(x) = (x - 1)(2x^2 + 3x + 1) \]

### Step 5: Factor the Quadratic
Now we need to factor \(2x^2 + 3x + 1\).

1. **Finding Factors**:
We need two numbers that multiply to \(2 \times 1 = 2\) and add to \(3\). The numbers \(2\) and \(1\) work:
\[ 2x^2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1) \]

### Step 6: Write the Complete Factorization
Combining all factors, we have:
\[ f(x) = (x - 1)(2x + 1)(x + 1) \]

### Final Answer
The complete factorization of the expression \(2x^3 + x^2 - 2x - 1\) is:
\[ (x - 1)(2x + 1)(x + 1) \]