Find the roots of the polynomial equation x4 + 2x3 + 3x² + 2x + 2 = 0

Toots. N given that - 1 + i is one of its

Given that \(-1 + i\) is a root of the polynomial \(x^4 + 2x^3 + 3x^2 + 2x + 2 = 0\), we can use the fact that for polynomials with real coefficients, the complex conjugate of any complex root must also be a root. Therefore, \(-1 - i\) is also a root.

### Step-by-Step Solution:

1. Since \(-1 + i\) and \(-1 - i\) are roots, we can find the quadratic factor they correspond to. The quadratic factor is given by:
   \[
   (x - (-1 + i))(x - (-1 - i)) = (x + 1 - i)(x + 1 + i)
   \]

2. Using the difference of squares formula:
   \[
   (x + 1)^2 - i^2 = (x + 1)^2 + 1 = x^2 + 2x + 2
   \]
   So, \((x + 1)^2 + 1 = x^2 + 2x + 2\).

3. Now, divide the given polynomial \(x^4 + 2x^3 + 3x^2 + 2x + 2\) by \(x^2 + 2x + 2\).

### Division Process:

We divide \(x^4 + 2x^3 + 3x^2 + 2x + 2\) by \(x^2 + 2x + 2\) using polynomial long division:

\[
\frac{x^4 + 2x^3 + 3x^2 + 2x + 2}{x^2 + 2x + 2}
\]

The first term is \(x^2\), so:
\[
x^2 \cdot (x^2 + 2x + 2) = x^4 + 2x^3 + 2x^2
\]
Subtracting:
\[
(x^4 + 2x^3 + 3x^2 + 2x + 2) - (x^4 + 2x^3 + 2x^2) = x^2 + 2x + 2
\]
Now, the next term is \(1\):
\[
1 \cdot (x^2 + 2x + 2) = x^2 + 2x + 2
\]
Subtracting:
\[
(x^2 + 2x + 2) - (x^2 + 2x + 2) = 0
\]

So, the quotient is \(x^2 + 1\) and the remainder is 0.

Thus, the given polynomial factors as:
\[
(x^2 + 2x + 2)(x^2 + 1) = 0
\]

### Solving for the Roots:

1. From \(x^2 + 2x + 2 = 0\), we already know that the roots are \(-1 + i\) and \(-1 - i\).

2. From \(x^2 + 1 = 0\), solving gives:
   \[
   x^2 = -1 \quad \Rightarrow \quad x = \pm i
   \]

### Final Roots:
The roots of the polynomial are:
\[
x = -1 + i, \quad x = -1 - i, \quad x = i, \quad x = -i
\]

Would you like further details on any step or have questions?

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Here are 5 related questions to explore further:

1. How do you factor polynomials with complex roots?
2. Can you use synthetic division to divide by a quadratic factor?
3. What is the significance of complex conjugate roots in polynomials?
4. How does the degree of a polynomial relate to the number of roots?
5. How can you graph a polynomial equation with complex roots?

**Tip:** For any polynomial with real coefficients, complex roots always come in conjugate pairs.

Find the roots of the polynomial equation x^4 + 2x^3 + 3x² + 2x + 2 = 0 given that -1 + i is one of its roots.

This detailed solution explains how to solve the polynomial equation x^4 + 2x^3 + 3x² + 2x + 2 = 0, factoring with complex roots such as -1 + i and -1 - i. The solution demonstrates polynomial division and solving quadratic factors, suitable for high school students in Grades 10-12.

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