Factorise: `x ^(3) - 3x ^(2) + 3x + 7`

Options

A. `(x-1) (x ^(2) - 4x + 7)`
B. `(x +1) (x ^(2) - 4 x + 7)`
C. `(x+1) (x ^(2) + 4x + 7)`
D. `(x -1) (x ^(2) + 4x + 7)`

Answer

To factorize the expression \( x^3 - 3x^2 + 3x + 7 \), we will use the hit and trial method to find a root, and then perform polynomial long division.

### Step-by-step Solution:

1. **Identify the polynomial**:
We start with the polynomial \( P(x) = x^3 - 3x^2 + 3x + 7 \).

2. **Use the hit and trial method**:
We will test some simple values for \( x \) to find a root.

- **Test \( x = 1 \)**:
\[ P(1) = 1^3 - 3(1^2) + 3(1) + 7 = 1 - 3 + 3 + 7 = 8 \quad (\text{not a root}) \]

- **Test \( x = -1 \)**:
\[ P(-1) = (-1)^3 - 3(-1)^2 + 3(-1) + 7 = -1 - 3 - 3 + 7 = 0 \quad (\text{is a root}) \]

Since \( x = -1 \) is a root, we can say that \( x + 1 \) is a factor of the polynomial.

3. **Perform polynomial long division**:
We will divide \( P(x) \) by \( x + 1 \).

- **Set up the division**:
\[ \text{Divide } x^3 - 3x^2 + 3x + 7 \text{ by } x + 1. \]

- **First term**:
Divide \( x^3 \) by \( x \) to get \( x^2 \).
\[ x^2 \cdot (x + 1) = x^3 + x^2 \]

- **Subtract**:
\[ (x^3 - 3x^2 + 3x + 7) - (x^3 + x^2) = -4x^2 + 3x + 7 \]

- **Next term**:
Divide \( -4x^2 \) by \( x \) to get \( -4x \).
\[ -4x \cdot (x + 1) = -4x^2 - 4x \]

- **Subtract**:
\[ (-4x^2 + 3x + 7) - (-4x^2 - 4x) = 7x + 7 \]

- **Final term**:
Divide \( 7x \) by \( x \) to get \( 7 \).
\[ 7 \cdot (x + 1) = 7x + 7 \]

- **Subtract**:
\[ (7x + 7) - (7x + 7) = 0 \]

Thus, the division gives us:
\[ P(x) = (x + 1)(x^2 - 4x + 7) \]

4. **Final Factorization**:
The polynomial \( x^3 - 3x^2 + 3x + 7 \) can be factored as:
\[ (x + 1)(x^2 - 4x + 7) \]