Factorise `x^(4) + x^(2) + 1 . `

Options

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

Explanation

To factorize the expression \( x^4 + x^2 + 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ x^4 + x^2 + 1 \]
We can recognize that this expression can be manipulated by introducing a new variable. Let \( y = x^2 \). Then, we can rewrite the expression as:
\[ y^2 + y + 1 \]

### Step 2: Factor the quadratic expression
Next, we will factor the quadratic expression \( y^2 + y + 1 \). To factor it, we can look for two numbers that multiply to \( 1 \) (the constant term) and add up to \( 1 \) (the coefficient of \( y \)). However, there are no such real numbers. Therefore, we can use the quadratic formula to find the roots:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = 1 \).

### Step 3: Calculate the discriminant
Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, this indicates that the quadratic does not factor over the real numbers.

### Step 4: Use complex roots
The roots of the quadratic are:
\[ y = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, we can express the quadratic in its factored form using these roots:
\[ y^2 + y + 1 = \left(y - \frac{-1 + i\sqrt{3}}{2}\right)\left(y - \frac{-1 - i\sqrt{3}}{2}\right) \]

### Step 5: Substitute back for \( y \)
Now substituting back \( y = x^2 \):
\[ x^4 + x^2 + 1 = \left(x^2 - \frac{-1 + i\sqrt{3}}{2}\right)\left(x^2 - \frac{-1 - i\sqrt{3}}{2}\right) \]

### Conclusion
Thus, the factorization of \( x^4 + x^2 + 1 \) over the complex numbers is:
\[ \left(x^2 + \frac{1}{2} - \frac{i\sqrt{3}}{2}\right)\left(x^2 + \frac{1}{2} + \frac{i\sqrt{3}}{2}\right) \]