Write the equation off lowest degree with real coefficients if two of its roots are -1 annd 1+i.

Options

A. `x^(3)+x^(2)+2=0`
B. `x^(3)-x^(2)-2=0`
C. `x^(3)-x+2=0`
D. `x^(3)-x^(2)+2=0`

Explanation

To find the equation of the lowest degree with real coefficients given the roots -1 and 1+i, we follow these steps:

### Step 1: Identify the roots
The roots given are:
- \( r_1 = -1 \)
- \( r_2 = 1 + i \)

Since complex roots occur in conjugate pairs, the conjugate of \( r_2 \) is also a root:
- \( r_3 = 1 - i \)

### Step 2: Write the factors from the roots
The polynomial can be expressed as a product of its factors based on its roots:
\[ (x - r_1)(x - r_2)(x - r_3) \]
Substituting the roots:
\[ (x + 1)(x - (1 + i))(x - (1 - i)) \]

### Step 3: Simplify the complex factors
First, simplify the complex factors:
\[ (x - (1 + i))(x - (1 - i)) = (x - 1 - i)(x - 1 + i) \]
This is a difference of squares:
\[ = (x - 1)^2 - i^2 \]
Since \( i^2 = -1 \):
\[ = (x - 1)^2 - (-1) = (x - 1)^2 + 1 \]

### Step 4: Expand \( (x - 1)^2 + 1 \)
Now, expand \( (x - 1)^2 \):
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Thus,
\[ (x - 1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2 \]

### Step 5: Combine with the other factor
Now, combine this with the factor from the root -1:
\[ (x + 1)(x^2 - 2x + 2) \]

### Step 6: Expand the product
Now, expand this product:
\[ = x(x^2 - 2x + 2) + 1(x^2 - 2x + 2) \]
\[ = x^3 - 2x^2 + 2x + x^2 - 2x + 2 \]
Combine like terms:
\[ = x^3 - 2x^2 + x^2 + 2x - 2x + 2 \]
\[ = x^3 - x^2 + 2 \]

### Final Equation
Thus, the equation of the lowest degree with real coefficients is:
\[ x^3 - x^2 + 2 = 0 \]