Find roots of polynomial $$ p(x) = 4x-36x^2+96x^3-80x^4 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \dfrac{ 1 }{ 2 }\\[1 em]x_3 &= \dfrac{ 1 }{ 5 }\\[1 em]x_4 &= \dfrac{ 1 }{ 2 } \end{aligned} $$Step 1:
Write polynomial in descending order
$$ \begin{aligned} 4x-36x^2+96x^3-80x^4 & = 0\\[1 em] -80x^4+96x^3-36x^2+4x & = 0 \end{aligned} $$Step 2:
Factor out $ \color{blue}{ -4x }$ from $ -80x^4+96x^3-36x^2+4x $ and solve two separate equations:
$$ \begin{aligned} -80x^4+96x^3-36x^2+4x & = 0\\[1 em] \color{blue}{ -4x }\cdot ( 20x^3-24x^2+9x-1 ) & = 0 \\[1 em] \color{blue}{ -4x = 0} ~~ \text{or} ~~ 20x^3-24x^2+9x-1 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 3:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 20x^3-24x^2+9x-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}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \pm \frac{ 1}{ 2} ~~ \pm \frac{ 1}{ 4} ~~ \pm \frac{ 1}{ 5} ~~ \pm \frac{ 1}{ 10} ~~ \pm \frac{ 1}{ 20} ~~ \end{aligned} $$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 }$
$$ \frac{ 20x^3-24x^2+9x-1}{ 2x-1} = 10x^2-7x+1 $$Step 4:
The next rational root is $ x = \dfrac{ 1 }{ 2 } $
$$ \frac{ 20x^3-24x^2+9x-1}{ 2x-1} = 10x^2-7x+1 $$Step 5:
The next rational root is $ x = \dfrac{ 1 }{ 5 } $
$$ \frac{ 10x^2-7x+1}{ 5x-1} = 2x-1 $$Step 6:
To find the last zero, solve equation $ 2x-1 = 0 $
$$ \begin{aligned} 2x-1 & = 0 \\[1 em] 2 \cdot x & = 1 \\[1 em] x & = \frac{ 1 }{ 2 } \end{aligned} $$