Find roots of polynomial $$ p(x) = s^6+2s^5+11s^4+20s^3+30s^2+25s+5 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -1\\[1 em]s_2 &= -0.2781\\[1 em]s_3 &= -0.5105+1.4297i\\[1 em]s_4 &= -0.5105-1.4297i\\[1 em]s_5 &= 0.1496+2.7888i\\[1 em]s_6 &= 0.1496-2.7888i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ s = -1 } $ is a root of polynomial $ s^6+2s^5+11s^4+20s^3+30s^2+25s+5 $.
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}{ 5 } $, with a single factor of 1 and 5.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \end{aligned} $$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}{ s+1 }$
$$ \frac{ s^6+2s^5+11s^4+20s^3+30s^2+25s+5}{ s+1} = s^5+s^4+10s^3+10s^2+20s+5 $$Step 2:
The next rational root is $ s = -1 $
$$ \frac{ s^6+2s^5+11s^4+20s^3+30s^2+25s+5}{ s+1} = s^5+s^4+10s^3+10s^2+20s+5 $$Step 3:
Polynomial $ s^5+s^4+10s^3+10s^2+20s+5 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.