Find roots of polynomial $$ p(x) = 3x^3-192 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 4\\[1 em]x_2 &= -2+2 \sqrt{ 3 }i\\[1 em]x_3 &= -2-2 \sqrt{ 3 }i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 4 } $ is a root of polynomial $ 3x^3-192 $.

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}{ 192 } $, with factors of 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96 and 192.

The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 192 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 48}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 96}{ 1} \pm \frac{ 192}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 16}{ 3} \pm \frac{ 24}{ 3} \pm \frac{ 32}{ 3} \pm \frac{ 48}{ 3} \pm \frac{ 64}{ 3} \pm \frac{ 96}{ 3} \pm \frac{ 192}{ 3} ~~ \end{aligned} $$

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( 4 \right) = 0 $ so $ x = 4 $ 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-4 }$

$$ \frac{ 3x^3-192}{ x-4} = 3x^2+12x+48 $$

Step 2:

The next rational root is $ x = 4 $

$$ \frac{ 3x^3-192}{ x-4} = 3x^2+12x+48 $$

Step 3:

The solutions of $ 3x^2+12x+48 = 0 $ are: $ x = -2+2 \sqrt{ 3 }i ~ \text{and} ~ x = -2-2 \sqrt{ 3 }i$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

Polynomial Roots Calculator