Find roots of polynomial $$ p(x) = 60x^8-13x^7-80x^6-1610x^5+371x^4+2163x^3-270x^2-540x-81 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 3\\[1 em]x_3 &= -1\\[1 em]x_4 &= -\frac{ 1 }{ 3 }\\[1 em]x_5 &= \frac{ 3 }{ 4 }\\[1 em]x_6 &= -\frac{ 1 }{ 5 }\\[1 em]x_7 &= -\frac{ 3 }{ 2 }+\frac{ 3 \sqrt{ 3}}{ 2 }i\\[1 em]x_8 &= -\frac{ 3 }{ 2 }-3 \frac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 60x^8-13x^7-80x^6-1610x^5+371x^4+2163x^3-270x^2-540x-81 $.
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}{ 81 } $, with factors of 1, 3, 9, 27 and 81.
The leading coefficient is $ \color{red}{ 60 }$, with factors of 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 81 }}{\text{ factors of 60 }} = \pm \dfrac{\text{ ( 1, 3, 9, 27, 81 ) }}{\text{ ( 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 81}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 27}{ 2} \pm \frac{ 81}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 27}{ 3} \pm \frac{ 81}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 27}{ 4} \pm \frac{ 81}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 9}{ 5} \pm \frac{ 27}{ 5} \pm \frac{ 81}{ 5} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 3}{ 6} \pm \frac{ 9}{ 6} \pm \frac{ 27}{ 6} \pm \frac{ 81}{ 6} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 9}{ 10} \pm \frac{ 27}{ 10} \pm \frac{ 81}{ 10} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 3}{ 12} \pm \frac{ 9}{ 12} \pm \frac{ 27}{ 12} \pm \frac{ 81}{ 12} ~~ \pm \frac{ 1}{ 15} \pm \frac{ 3}{ 15} \pm \frac{ 9}{ 15} \pm \frac{ 27}{ 15} \pm \frac{ 81}{ 15} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 3}{ 20} \pm \frac{ 9}{ 20} \pm \frac{ 27}{ 20} \pm \frac{ 81}{ 20} ~~ \pm \frac{ 1}{ 30} \pm \frac{ 3}{ 30} \pm \frac{ 9}{ 30} \pm \frac{ 27}{ 30} \pm \frac{ 81}{ 30} ~~ \pm \frac{ 1}{ 60} \pm \frac{ 3}{ 60} \pm \frac{ 9}{ 60} \pm \frac{ 27}{ 60} \pm \frac{ 81}{ 60} ~~ \end{aligned} $$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 }$
$$ \frac{ 60x^8-13x^7-80x^6-1610x^5+371x^4+2163x^3-270x^2-540x-81}{ x-1} = 60x^7+47x^6-33x^5-1643x^4-1272x^3+891x^2+621x+81 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 60x^8-13x^7-80x^6-1610x^5+371x^4+2163x^3-270x^2-540x-81}{ x-1} = 60x^7+47x^6-33x^5-1643x^4-1272x^3+891x^2+621x+81 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ 60x^7+47x^6-33x^5-1643x^4-1272x^3+891x^2+621x+81}{ x-3} = 60x^6+227x^5+648x^4+301x^3-369x^2-216x-27 $$Step 4:
The next rational root is $ x = -1 $
$$ \frac{ 60x^6+227x^5+648x^4+301x^3-369x^2-216x-27}{ x+1} = 60x^5+167x^4+481x^3-180x^2-189x-27 $$Step 5:
The next rational root is $ x = -\dfrac{ 1 }{ 3 } $
$$ \frac{ 60x^5+167x^4+481x^3-180x^2-189x-27}{ 3x+1} = 20x^4+49x^3+144x^2-108x-27 $$Step 6:
The next rational root is $ x = \dfrac{ 3 }{ 4 } $
$$ \frac{ 20x^4+49x^3+144x^2-108x-27}{ 4x-3} = 5x^3+16x^2+48x+9 $$Step 7:
The next rational root is $ x = -\dfrac{ 1 }{ 5 } $
$$ \frac{ 5x^3+16x^2+48x+9}{ 5x+1} = x^2+3x+9 $$Step 8:
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.