Find roots of polynomial $$ p(x) = x^3-\frac{105}{10}x^2+\frac{345}{10}x-35 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= 5\\[1 em]x_3 &= \dfrac{ 7 }{ 2 } \end{aligned} $$Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 10 } $.
$$ \begin{aligned} x^3-\frac{105}{10}x^2+\frac{345}{10}x-35 & = 0 ~~~ / \cdot \color{blue}{ 10 } \\[1 em] 10x^3-105x^2+345x-350 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 10x^3-105x^2+345x-350 $.
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}{ 350 } $, with factors of 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175 and 350.
The leading coefficient is $ \color{red}{ 10 }$, with factors of 1, 2, 5 and 10.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 350 }}{\text{ factors of 10 }} = \pm \dfrac{\text{ ( 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175, 350 ) }}{\text{ ( 1, 2, 5, 10 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 14}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 35}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 70}{ 1} \pm \frac{ 175}{ 1} \pm \frac{ 350}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 14}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 35}{ 2} \pm \frac{ 50}{ 2} \pm \frac{ 70}{ 2} \pm \frac{ 175}{ 2} \pm \frac{ 350}{ 2} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 5}{ 5} \pm \frac{ 7}{ 5} \pm \frac{ 10}{ 5} \pm \frac{ 14}{ 5} \pm \frac{ 25}{ 5} \pm \frac{ 35}{ 5} \pm \frac{ 50}{ 5} \pm \frac{ 70}{ 5} \pm \frac{ 175}{ 5} \pm \frac{ 350}{ 5}\\[ 1 em] \pm \frac{ 1}{ 10} & \pm \frac{ 2}{ 10} & \pm \frac{ 5}{ 10} & \pm \frac{ 7}{ 10} & \pm \frac{ 10}{ 10} & \pm \frac{ 14}{ 10} & \pm \frac{ 25}{ 10} & \pm \frac{ 35}{ 10} & \pm \frac{ 50}{ 10} & \pm \frac{ 70}{ 10} & \pm \frac{ 175}{ 10} & \pm \frac{ 350}{ 10} ~~ \end{aligned} $$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( 2 \right) = 0 $ so $ x = 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}{ x-2 }$
$$ \frac{ 10x^3-105x^2+345x-350}{ x-2} = 10x^2-85x+175 $$Step 3:
The next rational root is $ x = 2 $
$$ \frac{ 10x^3-105x^2+345x-350}{ x-2} = 10x^2-85x+175 $$Step 4:
The next rational root is $ x = 5 $
$$ \frac{ 10x^2-85x+175}{ x-5} = 10x-35 $$Step 5:
To find the last zero, solve equation $ 10x-35 = 0 $
$$ \begin{aligned} 10x-35 & = 0 \\[1 em] 10 \cdot x & = 35 \\[1 em] x & = \frac{ 35 }{ 10 } \\[1 em] x & = \frac{ 35 : 5 }{ 10 : 5} \\[1 em] x & = \frac{ 7 }{ 2 } \end{aligned} $$