Find roots of polynomial $$ p(x) = x^{20}-1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1\\[1 em]x_3 &= 0.5878+0.809i\\[1 em]x_4 &= 0.5878-0.809i\\[1 em]x_5 &= -0.5878+0.809i\\[1 em]x_6 &= -0.5878-0.809i\\[1 em]x_7 &= -0.309+0.9511i\\[1 em]x_8 &= -0.309-0.9511i\\[1 em]x_9 &= 0.809+0.5878i\\[1 em]x_{10} &= 0.809-0.5878i\\[1 em]x_{11} &= 0.309+0.9511i\\[1 em]x_{12} &= 0.309-0.9511i\\[1 em]x_{13} &= -0.809+0.5878i\\[1 em]x_{14} &= -0.809-0.5878i\\[1 em]x_{15} &= i\\[1 em]x_{16} &= -i\\[1 em]x_{17} &= 0.9511+0.309i\\[1 em]x_{18} &= 0.9511-0.309i\\[1 em]x_{19} &= -0.9511+0.309i\\[1 em]x_{20} &= -0.9511-0.309i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ x^{20}-1 $.
The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
The constant term is $ \color{blue}{ 1 } $, with a single factor of 1.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \end{aligned} $$Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
We can see that $ p\left( 1 \right) = 0 $ so $ x = 1 $ is a root of a polynomial $ p(x) $.
To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ x-1 }$
$$ \frac{ x^{20}-1}{ x-1} = x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ x^{20}-1}{ x-1} = x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1}{ x+1} = x^{18}+x^{16}+x^{14}+x^{12}+x^{10}+x^8+x^6+x^4+x^2+1 $$Step 4:
Polynomial $ x^{18}+x^{16}+x^{14}+x^{12}+x^{10}+x^8+x^6+x^4+x^2+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.