Find roots of polynomial $$ p(x) = 100x^4-80x^3-84x^2+80x-16 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1\\[1 em]x_3 &= \dfrac{ 2 }{ 5 }\\[1 em]x_4 &= \dfrac{ 2 }{ 5 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 100x^4-80x^3-84x^2+80x-16 $.
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}{ 16 } $, with factors of 1, 2, 4, 8 and 16.
The leading coefficient is $ \color{red}{ 100 }$, with factors of 1, 2, 4, 5, 10, 20, 25, 50 and 100.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 16 }}{\text{ factors of 100 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16 ) }}{\text{ ( 1, 2, 4, 5, 10, 20, 25, 50, 100 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 16}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 16}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 8}{ 5} \pm \frac{ 16}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 4}{ 10} \pm \frac{ 8}{ 10} \pm \frac{ 16}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 2}{ 20} \pm \frac{ 4}{ 20} \pm \frac{ 8}{ 20} \pm \frac{ 16}{ 20} ~~ \pm \frac{ 1}{ 25} \pm \frac{ 2}{ 25} \pm \frac{ 4}{ 25} \pm \frac{ 8}{ 25} \pm \frac{ 16}{ 25} ~~ \pm \frac{ 1}{ 50} \pm \frac{ 2}{ 50} \pm \frac{ 4}{ 50} \pm \frac{ 8}{ 50} \pm \frac{ 16}{ 50} ~~ \pm \frac{ 1}{ 100} \pm \frac{ 2}{ 100} \pm \frac{ 4}{ 100} \pm \frac{ 8}{ 100} \pm \frac{ 16}{ 100} ~~ \end{aligned} $$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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$
$$ \frac{ 100x^4-80x^3-84x^2+80x-16}{ x-1} = 100x^3+20x^2-64x+16 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 100x^4-80x^3-84x^2+80x-16}{ x-1} = 100x^3+20x^2-64x+16 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ 100x^3+20x^2-64x+16}{ x+1} = 100x^2-80x+16 $$Step 4:
The next rational root is $ x = \dfrac{ 2 }{ 5 } $
$$ \frac{ 100x^2-80x+16}{ 5x-2} = 20x-8 $$Step 5:
To find the last zero, solve equation $ 20x-8 = 0 $
$$ \begin{aligned} 20x-8 & = 0 \\[1 em] 20 \cdot x & = 8 \\[1 em] x & = \frac{ 8 }{ 20 } \\[1 em] x & = \frac{ 8 : 4 }{ 20 : 4} \\[1 em] x & = \frac{ 2 }{ 5 } \end{aligned} $$