Find roots of polynomial $$ p(x) = 3d^3-5d^2+2d-8 $$
Answer
The roots of polynomial $ p(d) $ are:
$$ \begin{aligned}d_1 &= 2\\[1 em]d_2 &= -\frac{ 1 }{ 6 }+\frac{\sqrt{ 47 }}{ 6 }i\\[1 em]d_3 &= -\frac{ 1 }{ 6 }- \frac{\sqrt{ 47 }}{ 6 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ d = 2 } $ is a root of polynomial $ 3d^3-5d^2+2d-8 $.
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}{ 8 } $, with factors of 1, 2, 4 and 8.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 8 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 8}{ 3} ~~ \end{aligned} $$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( 2 \right) = 0 $ so $ x = 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}{ d-2 }$
$$ \frac{ 3d^3-5d^2+2d-8}{ d-2} = 3d^2+d+4 $$Step 2:
The next rational root is $ d = 2 $
$$ \frac{ 3d^3-5d^2+2d-8}{ d-2} = 3d^2+d+4 $$Step 3:
The solutions of $ 3d^2+d+4 = 0 $ are: $ d = -\dfrac{ 1 }{ 6 }+\dfrac{\sqrt{ 47 }}{ 6 }i ~ \text{and} ~ d = -\dfrac{ 1 }{ 6 }-\dfrac{\sqrt{ 47 }}{ 6 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.