To find the zeros of the polynomial [tex]\( P(x) = x^3 - 2x^2 + 2x - 1 \)[/tex], we need to solve for [tex]\( x \)[/tex] such that [tex]\( P(x) = 0 \)[/tex].
Let's outline the steps involved:
1. Identify the Polynomial:
The polynomial given is [tex]\( P(x) = x^3 - 2x^2 + 2x - 1 \)[/tex].
2. Set the Polynomial Equal to Zero:
To find the roots, we need to solve:
[tex]\[
x^3 - 2x^2 + 2x - 1 = 0
\][/tex]
3. Find the Zeros:
By solving the equation [tex]\( x^3 - 2x^2 + 2x - 1 = 0 \)[/tex], we obtain the following roots:
[tex]\[
x = 1, \quad x = \frac{1}{2} - \frac{\sqrt{3}i}{2}, \quad x = \frac{1}{2} + \frac{\sqrt{3}i}{2}
\][/tex]
Therefore, the zeros of the polynomial [tex]\( P(x) = x^3 - 2x^2 + 2x - 1 \)[/tex] are [tex]\( 1, \frac{1}{2} - \frac{\sqrt{3}i}{2}, \)[/tex] and [tex]\( \frac{1}{2} + \frac{\sqrt{3}i}{2} \)[/tex].
These roots include one real root, [tex]\( x = 1 \)[/tex], and two complex conjugate roots, [tex]\( x = \frac{1}{2} - \frac{\sqrt{3}i}{2} \)[/tex] and [tex]\( x = \frac{1}{2} + \frac{\sqrt{3}i}{2} \)[/tex].
Hence, the zeros of the polynomial [tex]\( P(x) = x^3 - 2x^2 + 2x - 1 \)[/tex] are: [tex]\( 1, \frac{1}{2} - \frac{\sqrt{3}i}{2}, \frac{1}{2} + \frac{\sqrt{3}i}{2} \)[/tex].
