Find roots of polynomial $$ p(x) = m^4-2m^3+2m^2-2m+1 $$
Answer
The roots of polynomial $ p(m) $ are:
$$ \begin{aligned}m_1 &= 1\\[1 em]m_2 &= 1\\[1 em]m_3 &= i\\[1 em]m_4 &= -i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ m = 1 } $ is a root of polynomial $ m^4-2m^3+2m^2-2m+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}{ m-1 }$
$$ \frac{ m^4-2m^3+2m^2-2m+1}{ m-1} = m^3-m^2+m-1 $$Step 2:
The next rational root is $ m = 1 $
$$ \frac{ m^4-2m^3+2m^2-2m+1}{ m-1} = m^3-m^2+m-1 $$Step 3:
The next rational root is $ m = 1 $
$$ \frac{ m^3-m^2+m-1}{ m-1} = m^2+1 $$Step 4:
The solutions of $ m^2+1 = 0 $ are: $ m = i ~ \text{and} ~ m = -i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.