The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 2.1754\\[1 em]x_3 &= 0.5746 \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 8x^3-30x^2+32x-10 $.
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}{ 10 } $, with factors of 1, 2, 5 and 10.
The leading coefficient is $ \color{red}{ 8 }$, with factors of 1, 2, 4 and 8.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 10 }}{\text{ factors of 8 }} = \pm \dfrac{\text{ ( 1, 2, 5, 10 ) }}{\text{ ( 1, 2, 4, 8 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 10}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 10}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 10}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 2}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 10}{ 8}\\[ 1 em] \end{aligned} $$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 }$
$$ \frac{ 8x^3-30x^2+32x-10}{ x-1} = 8x^2-22x+10 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 8x^3-30x^2+32x-10}{ x-1} = 8x^2-22x+10 $$Step 3:
The solutions of $ 8x^2-22x+10 = 0 $ are: $ x = \dfrac{ 11 }{ 8 }-\dfrac{\sqrt{ 41 }}{ 8 } ~ \text{and} ~ x = \dfrac{ 11 }{ 8 }+\dfrac{\sqrt{ 41 }}{ 8 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.