Find roots of polynomial $$ p(x) = x^{16}-24x^{14}+228x^{12}-1144x^{10}+3342x^8-5832x^6+5940x^4-3240x^2+729 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 3\\[1 em]x_3 &= -1\\[1 em]x_4 &= -3\\[1 em]x_5 &= 1\\[1 em]x_6 &= -1\\[1 em]x_7 &= 1\\[1 em]x_8 &= -1\\[1 em]x_9 &= -1.7296\\[1 em]x_{10} &= -1.7345\\[1 em]x_{11} &= 1.7319+0.0001i\\[1 em]x_{12} &= 1.7319-0.0001i\\[1 em]x_{13} &= -1.7321+0.0025i\\[1 em]x_{14} &= -1.7321-0.0025i\\[1 em]x_{15} &= 1.7322+0.0002i\\[1 em]x_{16} &= 1.7322-0.0002i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ x^{16}-24x^{14}+228x^{12}-1144x^{10}+3342x^8-5832x^6+5940x^4-3240x^2+729 $.
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}{ 729 } $, with a single factor of 1, 3, 9, 27, 81, 243 and 729.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 729 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 9, 27, 81, 243, 729 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 243}{ 1} \pm \frac{ 729}{ 1} ~~ \end{aligned} $$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 }$
$$ \frac{ x^{16}-24x^{14}+228x^{12}-1144x^{10}+3342x^8-5832x^6+5940x^4-3240x^2+729}{ x-1} = x^{15}+x^{14}-23x^{13}-23x^{12}+205x^{11}+205x^{10}-939x^9-939x^8+2403x^7+2403x^6-3429x^5-3429x^4+2511x^3+2511x^2-729x-729 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ x^{16}-24x^{14}+228x^{12}-1144x^{10}+3342x^8-5832x^6+5940x^4-3240x^2+729}{ x-1} = x^{15}+x^{14}-23x^{13}-23x^{12}+205x^{11}+205x^{10}-939x^9-939x^8+2403x^7+2403x^6-3429x^5-3429x^4+2511x^3+2511x^2-729x-729 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ x^{15}+x^{14}-23x^{13}-23x^{12}+205x^{11}+205x^{10}-939x^9-939x^8+2403x^7+2403x^6-3429x^5-3429x^4+2511x^3+2511x^2-729x-729}{ x-3} = x^{14}+4x^{13}-11x^{12}-56x^{11}+37x^{10}+316x^9+9x^8-912x^7-333x^6+1404x^5+783x^4-1080x^3-729x^2+324x+243 $$Step 4:
The next rational root is $ x = -1 $
$$ \frac{ x^{14}+4x^{13}-11x^{12}-56x^{11}+37x^{10}+316x^9+9x^8-912x^7-333x^6+1404x^5+783x^4-1080x^3-729x^2+324x+243}{ x+1} = x^{13}+3x^{12}-14x^{11}-42x^{10}+79x^9+237x^8-228x^7-684x^6+351x^5+1053x^4-270x^3-810x^2+81x+243 $$Step 5:
The next rational root is $ x = -3 $
$$ \frac{ x^{13}+3x^{12}-14x^{11}-42x^{10}+79x^9+237x^8-228x^7-684x^6+351x^5+1053x^4-270x^3-810x^2+81x+243}{ x+3} = x^{12}-14x^{10}+79x^8-228x^6+351x^4-270x^2+81 $$Step 6:
The next rational root is $ x = 1 $
$$ \frac{ x^{12}-14x^{10}+79x^8-228x^6+351x^4-270x^2+81}{ x-1} = x^{11}+x^{10}-13x^9-13x^8+66x^7+66x^6-162x^5-162x^4+189x^3+189x^2-81x-81 $$Step 7:
The next rational root is $ x = -1 $
$$ \frac{ x^{11}+x^{10}-13x^9-13x^8+66x^7+66x^6-162x^5-162x^4+189x^3+189x^2-81x-81}{ x+1} = x^{10}-13x^8+66x^6-162x^4+189x^2-81 $$Step 8:
The next rational root is $ x = 1 $
$$ \frac{ x^{10}-13x^8+66x^6-162x^4+189x^2-81}{ x-1} = x^9+x^8-12x^7-12x^6+54x^5+54x^4-108x^3-108x^2+81x+81 $$Step 9:
The next rational root is $ x = -1 $
$$ \frac{ x^9+x^8-12x^7-12x^6+54x^5+54x^4-108x^3-108x^2+81x+81}{ x+1} = x^8-12x^6+54x^4-108x^2+81 $$Step 10:
Polynomial $ x^8-12x^6+54x^4-108x^2+81 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.