Find roots of polynomial $$ p(x) = 18+9x^5-11x^2-x^{13}+x^{14} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -1\\[1 em]x_2 &= -1.171\\[1 em]x_3 &= 0.966+0.5077i\\[1 em]x_4 &= 0.966-0.5077i\\[1 em]x_5 &= -0.9819+0.7903i\\[1 em]x_6 &= -0.9819-0.7903i\\[1 em]x_7 &= -0.4717+1.0552i\\[1 em]x_8 &= -0.4717-1.0552i\\[1 em]x_9 &= 0.7505+1.1484i\\[1 em]x_{10} &= 0.7505-1.1484i\\[1 em]x_{11} &= -0.0096+1.2975i\\[1 em]x_{12} &= -0.0096-1.2975i\\[1 em]x_{13} &= 1.332+0.381i\\[1 em]x_{14} &= 1.332-0.381i \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} 18+9x^5-11x^2-x^{13}+x^{14} & = 0\\[1 em] x^{14}-x^{13}+9x^5-11x^2+18 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -1 } $ is a root of polynomial $ x^{14}-x^{13}+9x^5-11x^2+18 $.
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}{ 18 } $, with a single factor of 1, 2, 3, 6, 9 and 18.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 18 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 9, 18 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 18}{ 1} ~~ \end{aligned} $$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 }$
$$ \frac{ x^{14}-x^{13}+9x^5-11x^2+18}{ x+1} = x^{13}-2x^{12}+2x^{11}-2x^{10}+2x^9-2x^8+2x^7-2x^6+2x^5+7x^4-7x^3+7x^2-18x+18 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ x^{14}-x^{13}+9x^5-11x^2+18}{ x+1} = x^{13}-2x^{12}+2x^{11}-2x^{10}+2x^9-2x^8+2x^7-2x^6+2x^5+7x^4-7x^3+7x^2-18x+18 $$Step 4:
Polynomial $ x^{13}-2x^{12}+2x^{11}-2x^{10}+2x^9-2x^8+2x^7-2x^6+2x^5+7x^4-7x^3+7x^2-18x+18 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.