Find roots of polynomial $$ p(x) = 4x^3-39x^2+92x-\frac{255}{4} $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 3 }{ 2 }\\[1 em]x_2 &= 6.653\\[1 em]x_3 &= 1.597 \end{aligned} $$Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 4 } $.
$$ \begin{aligned} 4x^3-39x^2+92x-\frac{255}{4} & = 0 ~~~ / \cdot \color{blue}{ 4 } \\[1 em] 16x^3-156x^2+368x-255 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 3 }{ 2 } } $ is a root of polynomial $ 16x^3-156x^2+368x-255 $.
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}{ 255 } $, with factors of 1, 3, 5, 15, 17, 51, 85 and 255.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 255 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1, 3, 5, 15, 17, 51, 85, 255 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 17}{ 1} \pm \frac{ 51}{ 1} \pm \frac{ 85}{ 1} \pm \frac{ 255}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 17}{ 2} \pm \frac{ 51}{ 2} \pm \frac{ 85}{ 2} \pm \frac{ 255}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 17}{ 4} \pm \frac{ 51}{ 4} \pm \frac{ 85}{ 4} \pm \frac{ 255}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 15}{ 8} \pm \frac{ 17}{ 8} \pm \frac{ 51}{ 8} \pm \frac{ 85}{ 8} \pm \frac{ 255}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 3}{ 16} \pm \frac{ 5}{ 16} \pm \frac{ 15}{ 16} \pm \frac{ 17}{ 16} \pm \frac{ 51}{ 16} \pm \frac{ 85}{ 16} \pm \frac{ 255}{ 16} ~~ \end{aligned} $$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{ 3 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 3 }{ 2 } $ 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}{ 2x-3 }$
$$ \frac{ 16x^3-156x^2+368x-255}{ 2x-3} = 8x^2-66x+85 $$Step 3:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ 16x^3-156x^2+368x-255}{ 2x-3} = 8x^2-66x+85 $$Step 4:
The solutions of $ 8x^2-66x+85 = 0 $ are: $ x = \dfrac{ 33 }{ 8 }-\dfrac{\sqrt{ 409 }}{ 8 } ~ \text{and} ~ x = \dfrac{ 33 }{ 8 }+\dfrac{\sqrt{ 409 }}{ 8 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.