Find roots of polynomial $$ p(x) = s^4-4s^3+6s^2-4s+1 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= 1\\[1 em]s_2 &= 1\\[1 em]s_3 &= 1\\[1 em]s_4 &= 1 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ s = 1 } $ is a root of polynomial $ s^4-4s^3+6s^2-4s+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}{ s-1 }$
$$ \frac{ s^4-4s^3+6s^2-4s+1}{ s-1} = s^3-3s^2+3s-1 $$Step 2:
The next rational root is $ s = 1 $
$$ \frac{ s^4-4s^3+6s^2-4s+1}{ s-1} = s^3-3s^2+3s-1 $$Step 3:
The next rational root is $ s = 1 $
$$ \frac{ s^3-3s^2+3s-1}{ s-1} = s^2-2s+1 $$Step 4:
The next rational root is $ s = 1 $
$$ \frac{ s^2-2s+1}{ s-1} = s-1 $$Step 5:
To find the last zero, solve equation $ s-1 = 0 $
$$ \begin{aligned} s-1 & = 0 \\[1 em] s & = 1 \end{aligned} $$