Find roots of polynomial $$ p(x) = 54x^6+45x^5-102x^4-69x^3+35x^2+16x-4 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 1 }{ 2 }\\[1 em]x_2 &= -\frac{ 2 }{ 3 }\\[1 em]x_3 &= -\frac{ 2 }{ 3 }\\[1 em]x_4 &= 0.2052\\[1 em]x_5 &= 1.1761\\[1 em]x_6 &= -1.3813 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 54x^6+45x^5-102x^4-69x^3+35x^2+16x-4 $.
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}{ 4 } $, with factors of 1, 2 and 4.
The leading coefficient is $ \color{red}{ 54 }$, with factors of 1, 2, 3, 6, 9, 18, 27 and 54.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 4 }}{\text{ factors of 54 }} = \pm \dfrac{\text{ ( 1, 2, 4 ) }}{\text{ ( 1, 2, 3, 6, 9, 18, 27, 54 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 4}{ 6} ~~ \pm \frac{ 1}{ 9} \pm \frac{ 2}{ 9} \pm \frac{ 4}{ 9} ~~ \pm \frac{ 1}{ 18} \pm \frac{ 2}{ 18} \pm \frac{ 4}{ 18} ~~ \pm \frac{ 1}{ 27} \pm \frac{ 2}{ 27} \pm \frac{ 4}{ 27} ~~ \pm \frac{ 1}{ 54} \pm \frac{ 2}{ 54} \pm \frac{ 4}{ 54} ~~ \end{aligned} $$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{ 1 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 2 } $ 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}{ 2x-1 }$
$$ \frac{ 54x^6+45x^5-102x^4-69x^3+35x^2+16x-4}{ 2x-1} = 27x^5+36x^4-33x^3-51x^2-8x+4 $$Step 2:
The next rational root is $ x = \dfrac{ 1 }{ 2 } $
$$ \frac{ 54x^6+45x^5-102x^4-69x^3+35x^2+16x-4}{ 2x-1} = 27x^5+36x^4-33x^3-51x^2-8x+4 $$Step 3:
The next rational root is $ x = -\dfrac{ 2 }{ 3 } $
$$ \frac{ 27x^5+36x^4-33x^3-51x^2-8x+4}{ 3x+2} = 9x^4+6x^3-15x^2-7x+2 $$Step 4:
The next rational root is $ x = -\dfrac{ 2 }{ 3 } $
$$ \frac{ 9x^4+6x^3-15x^2-7x+2}{ 3x+2} = 3x^3-5x+1 $$Step 5:
Polynomial $ 3x^3-5x+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.