Find roots of polynomial $$ p(x) = 9x^4-87x^3+240x^2-174x+120 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4\\[1 em]x_2 &= 5\\[1 em]x_3 &= \frac{ 1 }{ 3 }+\frac{\sqrt{ 5 }}{ 3 }i\\[1 em]x_4 &= \frac{ 1 }{ 3 }- \frac{\sqrt{ 5 }}{ 3 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 4 } $ is a root of polynomial $ 9x^4-87x^3+240x^2-174x+120 $.
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}{ 120 } $, with factors of 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
The leading coefficient is $ \color{red}{ 9 }$, with factors of 1, 3 and 9.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 120 }}{\text{ factors of 9 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ) }}{\text{ ( 1, 3, 9 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 40}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 120}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 10}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 15}{ 3} \pm \frac{ 20}{ 3} \pm \frac{ 24}{ 3} \pm \frac{ 30}{ 3} \pm \frac{ 40}{ 3} \pm \frac{ 60}{ 3} \pm \frac{ 120}{ 3} ~~ \pm \frac{ 1}{ 9} \pm \frac{ 2}{ 9} \pm \frac{ 3}{ 9} \pm \frac{ 4}{ 9} \pm \frac{ 5}{ 9} \pm \frac{ 6}{ 9} \pm \frac{ 8}{ 9} \pm \frac{ 10}{ 9} \pm \frac{ 12}{ 9} \pm \frac{ 15}{ 9} \pm \frac{ 20}{ 9} \pm \frac{ 24}{ 9} \pm \frac{ 30}{ 9} \pm \frac{ 40}{ 9} \pm \frac{ 60}{ 9} \pm \frac{ 120}{ 9} ~~ \end{aligned} $$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 }$
$$ \frac{ 9x^4-87x^3+240x^2-174x+120}{ x-4} = 9x^3-51x^2+36x-30 $$Step 2:
The next rational root is $ x = 4 $
$$ \frac{ 9x^4-87x^3+240x^2-174x+120}{ x-4} = 9x^3-51x^2+36x-30 $$Step 3:
The next rational root is $ x = 5 $
$$ \frac{ 9x^3-51x^2+36x-30}{ x-5} = 9x^2-6x+6 $$Step 4:
The solutions of $ 9x^2-6x+6 = 0 $ are: $ x = \dfrac{ 1 }{ 3 }+\dfrac{\sqrt{ 5 }}{ 3 }i ~ \text{and} ~ x = \dfrac{ 1 }{ 3 }-\dfrac{\sqrt{ 5 }}{ 3 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.