Find roots of polynomial $$ p(x) = x^3-27 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -\frac{ 3 }{ 2 }+\frac{ 3 \sqrt{ 3}}{ 2 }i\\[1 em]x_3 &= -\frac{ 3 }{ 2 }-3 \frac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^3-27 $.
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}{ 27 } $, with a single factor of 1, 3, 9 and 27.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 27 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 9, 27 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 27}{ 1} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ x^3-27}{ x-3} = x^2+3x+9 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ x^3-27}{ x-3} = x^2+3x+9 $$Step 3:
The solutions of $ x^2+3x+9 = 0 $ are: $ x = -\dfrac{ 3 }{ 2 }+\dfrac{ 3 \sqrt{ 3}}{ 2 }i ~ \text{and} ~ x = -\dfrac{ 3 }{ 2 }-\dfrac{ 3 \sqrt{ 3}}{ 2 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.