Find roots of polynomial $$ p(x) = \frac{12}{10}x^3-\frac{144}{100}x^2-\frac{6}{100}x+\frac{48}{100} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 1 }{ 2 }\\[1 em]x_2 &= \frac{ 17 }{ 20 }+\frac{\sqrt{ 31 }}{ 20 }i\\[1 em]x_3 &= \frac{ 17 }{ 20 }- \frac{\sqrt{ 31 }}{ 20 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} \frac{12}{10}x^3-\frac{144}{100}x^2-\frac{6}{100}x+\frac{48}{100} & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 120x^3-144x^2-6x+48 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 120x^3-144x^2-6x+48 $.
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}{ 48 } $, with factors of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
The leading coefficient is $ \color{red}{ 120 }$, with factors of 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 48 }}{\text{ factors of 120 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ) }}{\text{ ( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 48}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 24}{ 2} \pm \frac{ 48}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 16}{ 3} \pm \frac{ 24}{ 3} \pm \frac{ 48}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 12}{ 4} \pm \frac{ 16}{ 4} \pm \frac{ 24}{ 4} \pm \frac{ 48}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 6}{ 5} \pm \frac{ 8}{ 5} \pm \frac{ 12}{ 5} \pm \frac{ 16}{ 5} \pm \frac{ 24}{ 5} \pm \frac{ 48}{ 5} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 3}{ 6} \pm \frac{ 4}{ 6} \pm \frac{ 6}{ 6} \pm \frac{ 8}{ 6} \pm \frac{ 12}{ 6} \pm \frac{ 16}{ 6} \pm \frac{ 24}{ 6} \pm \frac{ 48}{ 6} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 2}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 4}{ 8} \pm \frac{ 6}{ 8} \pm \frac{ 8}{ 8} \pm \frac{ 12}{ 8} \pm \frac{ 16}{ 8} \pm \frac{ 24}{ 8} \pm \frac{ 48}{ 8} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 4}{ 10} \pm \frac{ 6}{ 10} \pm \frac{ 8}{ 10} \pm \frac{ 12}{ 10} \pm \frac{ 16}{ 10} \pm \frac{ 24}{ 10} \pm \frac{ 48}{ 10} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 2}{ 12} \pm \frac{ 3}{ 12} \pm \frac{ 4}{ 12} \pm \frac{ 6}{ 12} \pm \frac{ 8}{ 12} \pm \frac{ 12}{ 12} \pm \frac{ 16}{ 12} \pm \frac{ 24}{ 12} \pm \frac{ 48}{ 12} ~~ \pm \frac{ 1}{ 15} \pm \frac{ 2}{ 15} \pm \frac{ 3}{ 15} \pm \frac{ 4}{ 15} \pm \frac{ 6}{ 15} \pm \frac{ 8}{ 15} \pm \frac{ 12}{ 15} \pm \frac{ 16}{ 15} \pm \frac{ 24}{ 15} \pm \frac{ 48}{ 15} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 2}{ 20} \pm \frac{ 3}{ 20} \pm \frac{ 4}{ 20} \pm \frac{ 6}{ 20} \pm \frac{ 8}{ 20} \pm \frac{ 12}{ 20} \pm \frac{ 16}{ 20} \pm \frac{ 24}{ 20} \pm \frac{ 48}{ 20} ~~ \pm \frac{ 1}{ 24} \pm \frac{ 2}{ 24} \pm \frac{ 3}{ 24} \pm \frac{ 4}{ 24} \pm \frac{ 6}{ 24} \pm \frac{ 8}{ 24} \pm \frac{ 12}{ 24} \pm \frac{ 16}{ 24} \pm \frac{ 24}{ 24} \pm \frac{ 48}{ 24} ~~ \pm \frac{ 1}{ 30} \pm \frac{ 2}{ 30} \pm \frac{ 3}{ 30} \pm \frac{ 4}{ 30} \pm \frac{ 6}{ 30} \pm \frac{ 8}{ 30} \pm \frac{ 12}{ 30} \pm \frac{ 16}{ 30} \pm \frac{ 24}{ 30} \pm \frac{ 48}{ 30} ~~ \pm \frac{ 1}{ 40} \pm \frac{ 2}{ 40} \pm \frac{ 3}{ 40} \pm \frac{ 4}{ 40} \pm \frac{ 6}{ 40} \pm \frac{ 8}{ 40} \pm \frac{ 12}{ 40} \pm \frac{ 16}{ 40} \pm \frac{ 24}{ 40} \pm \frac{ 48}{ 40} ~~ \pm \frac{ 1}{ 60} \pm \frac{ 2}{ 60} \pm \frac{ 3}{ 60} \pm \frac{ 4}{ 60} \pm \frac{ 6}{ 60} \pm \frac{ 8}{ 60} \pm \frac{ 12}{ 60} \pm \frac{ 16}{ 60} \pm \frac{ 24}{ 60} \pm \frac{ 48}{ 60} ~~ \pm \frac{ 1}{ 120} \pm \frac{ 2}{ 120} \pm \frac{ 3}{ 120} \pm \frac{ 4}{ 120} \pm \frac{ 6}{ 120} \pm \frac{ 8}{ 120} \pm \frac{ 12}{ 120} \pm \frac{ 16}{ 120} \pm \frac{ 24}{ 120} \pm \frac{ 48}{ 120} ~~ \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 }$
$$ \frac{ 120x^3-144x^2-6x+48}{ 2x+1} = 60x^2-102x+48 $$Step 3:
The next rational root is $ x = -\dfrac{ 1 }{ 2 } $
$$ \frac{ 120x^3-144x^2-6x+48}{ 2x+1} = 60x^2-102x+48 $$Step 4:
The solutions of $ 60x^2-102x+48 = 0 $ are: $ x = \dfrac{ 17 }{ 20 }+\dfrac{\sqrt{ 31 }}{ 20 }i ~ \text{and} ~ x = \dfrac{ 17 }{ 20 }-\dfrac{\sqrt{ 31 }}{ 20 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.