Find roots of polynomial $$ p(x) = z^3-\frac{3}{4}z^2+\frac{1}{16} $$
The roots of polynomial $ p(z) $ are:
$$ \begin{aligned}z_1 &= \dfrac{ 1 }{ 2 }\\[1 em]z_2 &= -\dfrac{ 1 }{ 4 }\\[1 em]z_3 &= \dfrac{ 1 }{ 2 } \end{aligned} $$Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 16 } $.
$$ \begin{aligned} z^3-\frac{3}{4}z^2+\frac{1}{16} & = 0 ~~~ / \cdot \color{blue}{ 16 } \\[1 em] 16z^3-12z^2+1 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ z = \dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 16z^3-12z^2+1 $.
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}{ 1 } $, with factors of 1.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \pm \frac{ 1}{ 2} ~~ \pm \frac{ 1}{ 4} ~~ \pm \frac{ 1}{ 8} ~~ \pm \frac{ 1}{ 16} ~~ \end{aligned} $$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{ 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}{ 2z-1 }$
$$ \frac{ 16z^3-12z^2+1}{ 2z-1} = 8z^2-2z-1 $$Step 3:
The next rational root is $ z = \dfrac{ 1 }{ 2 } $
$$ \frac{ 16z^3-12z^2+1}{ 2z-1} = 8z^2-2z-1 $$Step 4:
The next rational root is $ z = -\dfrac{ 1 }{ 4 } $
$$ \frac{ 8z^2-2z-1}{ 4z+1} = 2z-1 $$Step 5:
To find the last zero, solve equation $ 2z-1 = 0 $
$$ \begin{aligned} 2z-1 & = 0 \\[1 em] 2 \cdot z & = 1 \\[1 em] z & = \frac{ 1 }{ 2 } \end{aligned} $$