The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 0.6076+0.992i\\[1 em]x_3 &= 0.6076-0.992i\\[1 em]x_4 &= -0.7843+0.9333i\\[1 em]x_5 &= -0.7843-0.9333i\\[1 em]x_6 &= 0.8407+0.3306i\\[1 em]x_7 &= 0.8407-0.3306i\\[1 em]x_8 &= -0.0521+1.1671i\\[1 em]x_9 &= -0.0521-1.1671i\\[1 em]x_{10} &= -1.1118+0.3216i\\[1 em]x_{11} &= -1.1118-0.3216i \end{aligned} $$Step 1:
Write polynomial in descending order
$$ \begin{aligned} 2x^3-x^2-7x+6+x^6+x^5-2x^{11} & = 0\\[1 em] -2x^{11}+x^6+x^5+2x^3-x^2-7x+6 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ -2x^{11}+x^6+x^5+2x^3-x^2-7x+6 $.
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}{ 6 } $, with factors of 1, 2, 3 and 6.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 6 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} ~~ \end{aligned} $$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 }$
$$ \frac{ -2x^{11}+x^6+x^5+2x^3-x^2-7x+6}{ x-1} = -2x^{10}-2x^9-2x^8-2x^7-2x^6-x^5+2x^2+x-6 $$Step 3:
The next rational root is $ x = 1 $
$$ \frac{ -2x^{11}+x^6+x^5+2x^3-x^2-7x+6}{ x-1} = -2x^{10}-2x^9-2x^8-2x^7-2x^6-x^5+2x^2+x-6 $$Step 4:
Polynomial $ -2x^{10}-2x^9-2x^8-2x^7-2x^6-x^5+2x^2+x-6 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.