Answer :
Sure, let's go through the problem step-by-step.
### (a) Finding all zeros of \( P(x) = x^6 - 1 \)
To find the zeros of \( P(x) = x^6 - 1 \), we need to solve the equation:
[tex]\[ x^6 - 1 = 0 \][/tex]
This can be rewritten as:
[tex]\[ x^6 = 1 \][/tex]
The solutions to this equation are the sixth roots of unity. These roots are given by:
[tex]\[ x = e^{2\pi ik/6} \][/tex]
for \( k = 0, 1, 2, 3, 4, 5 \).
Expressing these roots in terms of real and imaginary components, we get:
- For \( k = 0 \):
[tex]\[ x = e^{0} = 1 \][/tex]
- For \( k = 1 \):
[tex]\[ x = e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 2 \):
[tex]\[ x = e^{2i\pi/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 3 \):
[tex]\[ x = e^{i\pi} = -1 \][/tex]
- For \( k = 4 \):
[tex]\[ x = e^{4i\pi/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 5 \):
[tex]\[ x = e^{5i\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i \][/tex]
Therefore, the zeros (roots) of \( P(x) \) are:
[tex]\[ \{-1, 1, -\frac{1}{2} - \frac{\sqrt{3}}{2}i, -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i\} \][/tex]
### (b) Factoring \( P(x) \)
The polynomial \( P(x) = x^6 - 1 \) can be factored into linear factors corresponding to the roots, as follows:
[tex]\[ x^6 - 1 = (x - 1)(x + 1)\left(x - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\right) \][/tex]
However, this factorization can be simplified by recognizing that the polynomial \( P(x) = x^6 - 1 \) can also be factored using its quadratic factors:
[tex]\[ P(x) = (x^2 - 1)\left((x + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right)\left((x - \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right) \][/tex]
Simplifying, we have:
[tex]\[ P(x) = (x^2 - 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
So, the complete factorization of \( P(x) \) is:
[tex]\[ P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
### Final Answer:
(a) The zeros of \( P(x) = x^6 - 1 \) are:
[tex]\[ x = -1, 1, -\frac{1}{2} - \frac{\sqrt{3}}{2}i, -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
(b) The complete factorization of \( P(x) = x^6 - 1 \) is:
[tex]\[ P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
### (a) Finding all zeros of \( P(x) = x^6 - 1 \)
To find the zeros of \( P(x) = x^6 - 1 \), we need to solve the equation:
[tex]\[ x^6 - 1 = 0 \][/tex]
This can be rewritten as:
[tex]\[ x^6 = 1 \][/tex]
The solutions to this equation are the sixth roots of unity. These roots are given by:
[tex]\[ x = e^{2\pi ik/6} \][/tex]
for \( k = 0, 1, 2, 3, 4, 5 \).
Expressing these roots in terms of real and imaginary components, we get:
- For \( k = 0 \):
[tex]\[ x = e^{0} = 1 \][/tex]
- For \( k = 1 \):
[tex]\[ x = e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 2 \):
[tex]\[ x = e^{2i\pi/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 3 \):
[tex]\[ x = e^{i\pi} = -1 \][/tex]
- For \( k = 4 \):
[tex]\[ x = e^{4i\pi/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \][/tex]
- For \( k = 5 \):
[tex]\[ x = e^{5i\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i \][/tex]
Therefore, the zeros (roots) of \( P(x) \) are:
[tex]\[ \{-1, 1, -\frac{1}{2} - \frac{\sqrt{3}}{2}i, -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i\} \][/tex]
### (b) Factoring \( P(x) \)
The polynomial \( P(x) = x^6 - 1 \) can be factored into linear factors corresponding to the roots, as follows:
[tex]\[ x^6 - 1 = (x - 1)(x + 1)\left(x - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\right)\left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\right) \][/tex]
However, this factorization can be simplified by recognizing that the polynomial \( P(x) = x^6 - 1 \) can also be factored using its quadratic factors:
[tex]\[ P(x) = (x^2 - 1)\left((x + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right)\left((x - \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right) \][/tex]
Simplifying, we have:
[tex]\[ P(x) = (x^2 - 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
So, the complete factorization of \( P(x) \) is:
[tex]\[ P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
### Final Answer:
(a) The zeros of \( P(x) = x^6 - 1 \) are:
[tex]\[ x = -1, 1, -\frac{1}{2} - \frac{\sqrt{3}}{2}i, -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \frac{1}{2} - \frac{\sqrt{3}}{2}i, \frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
(b) The complete factorization of \( P(x) = x^6 - 1 \) is:
[tex]\[ P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \][/tex]
