Find roots of polynomial $$ p(x) = 98x^3-133x^2+52x-5 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 1 }{ 2 }\\[1 em]x_2 &= \dfrac{ 1 }{ 7 }\\[1 em]x_3 &= \dfrac{ 5 }{ 7 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 98x^3-133x^2+52x-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}{ 98 }$, with factors of 1, 2, 7, 14, 49 and 98.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 98 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1, 2, 7, 14, 49, 98 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} ~~ \pm \frac{ 1}{ 7} \pm \frac{ 5}{ 7} ~~ \pm \frac{ 1}{ 14} \pm \frac{ 5}{ 14} ~~ \pm \frac{ 1}{ 49} \pm \frac{ 5}{ 49} ~~ \pm \frac{ 1}{ 98} \pm \frac{ 5}{ 98} ~~ \end{aligned} $$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 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 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-1 }$
$$ \frac{ 98x^3-133x^2+52x-5}{ 2x-1} = 49x^2-42x+5 $$Step 2:
The next rational root is $ x = \dfrac{ 1 }{ 2 } $
$$ \frac{ 98x^3-133x^2+52x-5}{ 2x-1} = 49x^2-42x+5 $$Step 3:
The next rational root is $ x = \dfrac{ 1 }{ 7 } $
$$ \frac{ 49x^2-42x+5}{ 7x-1} = 7x-5 $$Step 4:
To find the last zero, solve equation $ 7x-5 = 0 $
$$ \begin{aligned} 7x-5 & = 0 \\[1 em] 7 \cdot x & = 5 \\[1 em] x & = \frac{ 5 }{ 7 } \end{aligned} $$