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.