To find a polynomial equation with real coefficients that has the given zeros [tex]\( 1 - 2i \)[/tex] and [tex]\( 1 + 2i \)[/tex], follow these steps:
1. Understanding Conjugate Pairs:
- Given the roots [tex]\(1 - 2i\)[/tex] and [tex]\(1 + 2i\)[/tex], note that complex roots always occur in conjugate pairs when the polynomial has real coefficients.
2. Form the Factored Polynomial:
- We can express the polynomial as the product of two binomials, each reflecting one of the roots.
[tex]\[
f(x) = (x - (1 - 2i))(x - (1 + 2i))
\][/tex]
3. Simplify the Expression:
- Substitute the roots into the polynomial:
[tex]\[
f(x) = (x - 1 + 2i)(x - 1 - 2i)
\][/tex]
4. Using the Difference of Squares Formula:
- Recognize this as a difference of squares situation:
[tex]\[
(a + bi)(a - bi) = a^2 - (bi)^2
\][/tex]
- Here, [tex]\( a = x - 1 \)[/tex] and [tex]\( b = 2i \)[/tex].
5. Apply the Formula:
- Compute [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = (x - 1)^2 - (2i)^2
\][/tex]
- Simplify [tex]\( (2i)^2 \)[/tex]:
[tex]\[
(2i)^2 = 4i^2 = 4(-1) = -4
\][/tex]
- Substituting this back, we get:
[tex]\[
f(x) = (x - 1)^2 - (-4)
\][/tex]
[tex]\[
f(x) = (x - 1)^2 + 4
\][/tex]
6. Expand [tex]\((x - 1)^2\)[/tex]:
- Compute [tex]\((x - 1)^2\)[/tex]:
[tex]\[
(x - 1)^2 = x^2 - 2x + 1
\][/tex]
7. Combine Terms:
- Add the constant term [tex]\(4\)[/tex]:
[tex]\[
f(x) = x^2 - 2x + 1 + 4
\][/tex]
- Simplify the polynomial:
[tex]\[
f(x) = x^2 - 2x + 5
\][/tex]
So the polynomial equation with real coefficients that has the given zeros [tex]\( 1 - 2i \)[/tex] and [tex]\( 1 + 2i \)[/tex] is:
[tex]\[
x^2 - 2x + 5 = 0
\][/tex]
Thus, the completed equation is:
[tex]\[
x^2 - 2x + 5 = 0
\][/tex]
