Find roots of polynomial $$ p(x) = \frac{1}{5}x^4-\frac{4}{5}x^3+\frac{12}{5}x^2+\frac{4}{5}x-\frac{13}{5} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1\\[1 em]x_3 &= 2+3i\\[1 em]x_4 &= 2-3i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 5 } $.
$$ \begin{aligned} \frac{1}{5}x^4-\frac{4}{5}x^3+\frac{12}{5}x^2+\frac{4}{5}x-\frac{13}{5} & = 0 ~~~ / \cdot \color{blue}{ 5 } \\[1 em] x^4-4x^3+12x^2+4x-13 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ x^4-4x^3+12x^2+4x-13 $.
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}{ 13 } $, with a single factor of 1 and 13.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 13 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 13 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 13}{ 1} ~~ \end{aligned} $$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 }$
$$ \frac{ x^4-4x^3+12x^2+4x-13}{ x-1} = x^3-3x^2+9x+13 $$Step 3:
The next rational root is $ x = 1 $
$$ \frac{ x^4-4x^3+12x^2+4x-13}{ x-1} = x^3-3x^2+9x+13 $$Step 4:
The next rational root is $ x = -1 $
$$ \frac{ x^3-3x^2+9x+13}{ x+1} = x^2-4x+13 $$Step 5:
The solutions of $ x^2-4x+13 = 0 $ are: $ x = 2+3i ~ \text{and} ~ x = 2-3i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.