The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 1 }{ 3 }\\[1 em]x_2 &= -\frac{ 5 }{ 17 }\\[1 em]x_3 &= i\\[1 em]x_4 &= -i \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 1 }{ 3 } } $ is a root of polynomial $ 51x^4-2x^3+46x^2-2x-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 factors of 1 and 5.
The leading coefficient is $ \color{red}{ 51 }$, with factors of 1, 3, 17 and 51.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 51 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1, 3, 17, 51 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 5}{ 3} ~~ \pm \frac{ 1}{ 17} \pm \frac{ 5}{ 17} ~~ \pm \frac{ 1}{ 51} \pm \frac{ 5}{ 51}\\[ 1 em] \end{aligned} $$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( \dfrac{ 1 }{ 3 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 3 } $ 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}{ 3x-1 }$
$$ \frac{ 51x^4-2x^3+46x^2-2x-5}{ 3x-1} = 17x^3+5x^2+17x+5 $$Step 2:
The next rational root is $ x = \dfrac{ 1 }{ 3 } $
$$ \frac{ 51x^4-2x^3+46x^2-2x-5}{ 3x-1} = 17x^3+5x^2+17x+5 $$Step 3:
The next rational root is $ x = -\dfrac{ 5 }{ 17 } $
$$ \frac{ 17x^3+5x^2+17x+5}{ 17x+5} = x^2+1 $$Step 4:
The solutions of $ x^2+1 = 0 $ are: $ x = i ~ \text{and} ~ x = -i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.