Find roots of polynomial $$ p(x) = 10x^3+32x^2+120x+384 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\dfrac{ 16 }{ 5 }\\[1 em]x_2 &= 2 \sqrt{ 3 }i\\[1 em]x_3 &= -2 \sqrt{ 3 }i \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 16 }{ 5 } } $ is a root of polynomial $ 10x^3+32x^2+120x+384 $.
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}{ 384 } $, with factors of 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192 and 384.
The leading coefficient is $ \color{red}{ 10 }$, with factors of 1, 2, 5 and 10.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 384 }}{\text{ factors of 10 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 384 ) }}{\text{ ( 1, 2, 5, 10 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 48}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 96}{ 1} \pm \frac{ 128}{ 1} \pm \frac{ 192}{ 1} \pm \frac{ 384}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 24}{ 2} \pm \frac{ 32}{ 2} \pm \frac{ 48}{ 2} \pm \frac{ 64}{ 2} \pm \frac{ 96}{ 2} \pm \frac{ 128}{ 2} \pm \frac{ 192}{ 2} \pm \frac{ 384}{ 2} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 6}{ 5} \pm \frac{ 8}{ 5} \pm \frac{ 12}{ 5} \pm \frac{ 16}{ 5} \pm \frac{ 24}{ 5} \pm \frac{ 32}{ 5} \pm \frac{ 48}{ 5} \pm \frac{ 64}{ 5} \pm \frac{ 96}{ 5} \pm \frac{ 128}{ 5} \pm \frac{ 192}{ 5} \pm \frac{ 384}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 4}{ 10} \pm \frac{ 6}{ 10} \pm \frac{ 8}{ 10} \pm \frac{ 12}{ 10} \pm \frac{ 16}{ 10} \pm \frac{ 24}{ 10} \pm \frac{ 32}{ 10} \pm \frac{ 48}{ 10} \pm \frac{ 64}{ 10} \pm \frac{ 96}{ 10} \pm \frac{ 128}{ 10} \pm \frac{ 192}{ 10} \pm \frac{ 384}{ 10}\\[ 1 em] \end{aligned} $$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( -\dfrac{ 16 }{ 5 } \right) = 0 $ so $ x = -\dfrac{ 16 }{ 5 } $ 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}{ 5x+16 }$
$$ \frac{ 10x^3+32x^2+120x+384}{ 5x+16} = 2x^2+24 $$Step 2:
The next rational root is $ x = -\dfrac{ 16 }{ 5 } $
$$ \frac{ 10x^3+32x^2+120x+384}{ 5x+16} = 2x^2+24 $$Step 3:
The solutions of $ 2x^2+24 = 0 $ are: $ x = 2 \sqrt{ 3 } i ~ \text{and} ~ x = -2 \sqrt{ 3 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.