factor the following and show your solution.   1. x² - x + 2 = 0 2. 3x² - x + 12 = 0 3. x² + 2x = 0 4. 2x² - 4x + 2 = 0

Let's solve each of these quadratic equations by factoring them step by step.

### 1. \( x^2 - x + 2 = 0 \)

This quadratic equation does not factor nicely into rational numbers. To solve it, we'll use the quadratic formula:

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

For the equation \( x^2 - x + 2 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = -1 \)
- \( c = 2 \)

Substituting into the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)}
\]
\[
x = \frac{1 \pm \sqrt{1 - 8}}{2}
\]
\[
x = \frac{1 \pm \sqrt{-7}}{2}
\]

The solutions are complex:
\[
x = \frac{1 \pm \sqrt{7}i}{2}
\]

So the factorization in terms of complex numbers is:
\[
(x - \frac{1 + \sqrt{7}i}{2})(x - \frac{1 - \sqrt{7}i}{2})
\]

### 2. \( 3x^2 - x + 12 = 0 \)

Again, this quadratic equation does not factor nicely. Let's apply the quadratic formula:

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

Here:
- \( a = 3 \)
- \( b = -1 \)
- \( c = 12 \)

Substituting:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(12)}}{2(3)}
\]
\[
x = \frac{1 \pm \sqrt{1 - 144}}{6}
\]
\[
x = \frac{1 \pm \sqrt{-143}}{6}
\]

The solutions are complex:
\[
x = \frac{1 \pm \sqrt{143}i}{6}
\]

So the factorization is:
\[
\left(x - \frac{1 + \sqrt{143}i}{6}\right)\left(x - \frac{1 - \sqrt{143}i}{6}\right)
\]

### 3. \( x^2 + 2x = 0 \)

We can factor this quadratic equation easily by factoring out the common term \(x\):

\[
x(x + 2) = 0
\]

Setting each factor to zero gives the solutions:
\[
x = 0 \quad \text{or} \quad x = -2
\]

So the factorization is:
\[
x(x + 2) = 0
\]

### 4. \( 2x^2 - 4x + 2 = 0 \)

First, factor out the common factor of 2:

\[
2(x^2 - 2x + 1) = 0
\]

Notice that \(x^2 - 2x + 1\) is a perfect square trinomial:

\[
2(x - 1)^2 = 0
\]

Setting the expression inside the parentheses to zero gives:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

So the factorization is:
\[
2(x - 1)^2 = 0
\]

---

Would you like further details on any of these solutions? Here are some related questions you might consider:

1. How do you solve quadratic equations using the quadratic formula?
2. What are the conditions for a quadratic equation to have complex roots?
3. How can you identify perfect square trinomials?
4. What is the importance of factoring in solving quadratic equations?
5. How does completing the square relate to factoring?

**Tip:** When factoring, always look for common factors first and check if the quadratic can be rewritten as a perfect square trinomial.

Explore solutions to quadratic equations including x² - x + 2 = 0, 3x² - x + 12 = 0, x² + 2x = 0, and 2x² - 4x + 2 = 0 with detailed step-by-step explanations and factorization methods. Learn about complex roots, factoring techniques, and the quadratic formula. Suitable for high school students.

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