Given that \(-1 + 2i\) is a root of the polynomial equation \(x^4 - 12x + 5 = 0\), we can use the fact that the coefficients of the polynomial are real numbers. Therefore, the complex conjugate of the root, \(-1 - 2i\), must also be a root.

Let's denote the roots by \(r_1 = -1 + 2i\), \(r_2 = -1 - 2i\), and let \(r_3\) and \(r_4\) be the other two roots of the polynomial.

The polynomial can then be factored as:
\[
(x - r_1)(x - r_2)(x - r_3)(x - r_4) = 0
\]

First, let's find the quadratic factor corresponding to the complex roots \(r_1\) and \(r_2\):
\[
(x - (-1 + 2i))(x - (-1 - 2i))
\]

This simplifies to:
\[
(x + 1 - 2i)(x + 1 + 2i)
\]

This can be expanded using the difference of squares:
\[
(x + 1)^2 - (2i)^2 = (x + 1)^2 - 4(-1) = (x + 1)^2 + 4
\]

Expanding \((x + 1)^2 + 4\):
\[
(x^2 + 2x + 1) + 4 = x^2 + 2x + 5
\]

So, the quadratic factor is \(x^2 + 2x + 5\).

Now, the polynomial \(x^4 - 12x + 5\) can be factored as:
\[
(x^2 + 2x + 5)(x^2 + ax + b) = x^4 - 12x + 5
\]

We need to expand and match coefficients to find \(a\) and \(b\).

Expanding the product:
\[
(x^2 + 2x + 5)(x^2 + ax + b) = x^4 + ax^3 + bx^2 + 2x^3 + 2ax^2 + 2bx + 5x^2 + 5ax + 5b
\]
\[
= x^4 + (a + 2)x^3 + (b + 2a + 5)x^2 + (2b + 5a)x + 5b
\]

We need to match this with the original polynomial \(x^4 - 12x + 5\). The coefficients are:
\[
\text{Coefficient of } x^4: 1
\]
\[
\text{Coefficient of } x^3: 0 = a + 2 \quad \Rightarrow \quad a = -2
\]
\[
\text{Coefficient of } x^2: 0 = b + 2a + 5 = b + 2(-2) + 5 = b - 4 + 5 = b + 1 \quad \Rightarrow \quad b = -1
\]
\[
\text{Coefficient of } x: -12 = 2b + 5a = 2(-1) + 5(-2) = -2 - 10 = -12 \quad \text{(Matches)}
\]
\[
\text{Constant term: } 5 = 5b = 5(-1) = -5 \quad \text{(Matches)}
\]

Thus, the quadratic factor is \(x^2 - 2x - 1\).

The fully factored form of the polynomial is:
\[
(x^2 + 2x + 5)(x^2 - 2x - 1) = 0
\]

The roots are \(-1 + 2i\), \(-1 - 2i\), and the roots of \(x^2 - 2x - 1 = 0\), which are:
\[
x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}
\]

So, the roots of the polynomial are:
\[
\boxed{-1 + 2i, -1 - 2i, 1 + \sqrt{2}, 1 - \sqrt{2}}
\]

Would you like further details or have any other questions?

### Relative Questions
1. How do you factor polynomials with complex roots in general?
2. What is the significance of the complex conjugate root theorem?
3. How can you determine the number of real and complex roots of a polynomial?
4. How can the discriminant of a quadratic equation help determine the nature of its roots?
5. How do you use the quadratic formula to solve quadratic equations with irrational roots?

### Tip
When dealing with polynomials with real coefficients, always remember that complex roots occur in conjugate pairs.

Learn how to solve the polynomial equation x^4 - 12x + 5 = 0 given that -1 + 2i is a root. Understand the application of the complex conjugate root theorem and factorization of polynomials. Suitable for advanced high school students.

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