Find roots of polynomial $$ p(x) = 27x^6+26x^3-1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -1\\[1 em]x_2 &= \frac{ 1 }{ 3 }\\[1 em]x_3 &= -0.1667+0.2887i\\[1 em]x_4 &= -0.1667-0.2887i\\[1 em]x_5 &= 0.5+0.866i\\[1 em]x_6 &= 0.5-0.866i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -1 } $ is a root of polynomial $ 27x^6+26x^3-1 $.
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}{ 27 }$, with factors of 1, 3, 9 and 27.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 27 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 3, 9, 27 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \pm \frac{ 1}{ 3} ~~ \pm \frac{ 1}{ 9} ~~ \pm \frac{ 1}{ 27}\\[ 1 em] \end{aligned} $$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 }$
$$ \frac{ 27x^6+26x^3-1}{ x+1} = 27x^5-27x^4+27x^3-x^2+x-1 $$Step 2:
The next rational root is $ x = -1 $
$$ \frac{ 27x^6+26x^3-1}{ x+1} = 27x^5-27x^4+27x^3-x^2+x-1 $$Step 3:
The next rational root is $ x = \dfrac{ 1 }{ 3 } $
$$ \frac{ 27x^5-27x^4+27x^3-x^2+x-1}{ 3x-1} = 9x^4-6x^3+7x^2+2x+1 $$Step 4:
Polynomial $ 9x^4-6x^3+7x^2+2x+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.