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The problem asks to **factor** the quadratic expression \( x^2 - 4x + 5 \) over the set of complex numbers.

### Step-by-Step Solution:
We are given the quadratic \( x^2 - 4x + 5 \). To factor it, we need to solve for the roots using the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( x^2 - 4x + 5 \), we identify:
- \( a = 1 \)
- \( b = -4 \)
- \( c = 5 \)

Now, calculate the discriminant \( \Delta = b^2 - 4ac \):

\[
\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4
\]

Since the discriminant is negative, the roots will be complex. Now apply the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{-4}}{2(1)} = \frac{4 \pm \sqrt{-4}}{2}
\]

Since \( \sqrt{-4} = 2i \), we have:

\[
x = \frac{4 \pm 2i}{2} = 2 \pm i
\]

Thus, the two roots are \( x = 2 + i \) and \( x = 2 - i \).

### Factored Form:
Using these roots, the factored form is:

\[
(x - (2 + i))(x - (2 - i))
\]

Alternatively, this can be expressed as:

\[
(x - 2 - i)(x - 2 + i)
\]

Would you like further details or have any other questions?

### Related Questions:
1. What are the conditions for a quadratic equation to have complex roots?
2. How can you factor quadratics when the discriminant is negative?
3. What does it mean for a number to be "complex"?
4. How do you complete the square to solve quadratic equations?
5. Can all polynomials be factored over the complex numbers?

### Tip:
When factoring quadratic equations with a negative discriminant, always expect complex conjugate roots.

Factor x^2 - 4x + 5 over the set of complex numbers.

Learn how to factor the quadratic equation x^2 - 4x + 5 over the set of complex numbers. This problem demonstrates key algebraic concepts like the quadratic formula, solving for complex roots, and factoring quadratic equations with negative discriminants.

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