Find roots of polynomial $$ p(x) = 3x^4+5x^3+81x+135 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -3\\[1 em]x_2 &= -\frac{ 5 }{ 3 }\\[1 em]x_3 &= \frac{ 3 }{ 2 }+\frac{ 3 \sqrt{ 3}}{ 2 }i\\[1 em]x_4 &= \frac{ 3 }{ 2 }-3 \frac{\sqrt{ 3 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -3 } $ is a root of polynomial $ 3x^4+5x^3+81x+135 $.
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}{ 135 } $, with factors of 1, 3, 5, 9, 15, 27, 45 and 135.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 135 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 27, 45, 135 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 135}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 15}{ 3} \pm \frac{ 27}{ 3} \pm \frac{ 45}{ 3} \pm \frac{ 135}{ 3} ~~ \end{aligned} $$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( -3 \right) = 0 $ so $ x = -3 $ 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+3 }$
$$ \frac{ 3x^4+5x^3+81x+135}{ x+3} = 3x^3-4x^2+12x+45 $$Step 2:
The next rational root is $ x = -3 $
$$ \frac{ 3x^4+5x^3+81x+135}{ x+3} = 3x^3-4x^2+12x+45 $$Step 3:
The next rational root is $ x = -\dfrac{ 5 }{ 3 } $
$$ \frac{ 3x^3-4x^2+12x+45}{ 3x+5} = x^2-3x+9 $$Step 4:
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.