Find roots of polynomial $$ p(x) = 3x^5+15x^4+21x^3-15x^2-132x-180 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= -2\\[1 em]x_3 &= -3\\[1 em]x_4 &= -1+2i\\[1 em]x_5 &= -1-2i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 3x^5+15x^4+21x^3-15x^2-132x-180 $.
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}{ 180 } $, with factors of 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90 and 180.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 180 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 90}{ 1} \pm \frac{ 180}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 10}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 15}{ 3} \pm \frac{ 18}{ 3} \pm \frac{ 20}{ 3} \pm \frac{ 30}{ 3} \pm \frac{ 36}{ 3} \pm \frac{ 45}{ 3} \pm \frac{ 60}{ 3} \pm \frac{ 90}{ 3} \pm \frac{ 180}{ 3} ~~ \end{aligned} $$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}{ x-2 }$
$$ \frac{ 3x^5+15x^4+21x^3-15x^2-132x-180}{ x-2} = 3x^4+21x^3+63x^2+111x+90 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ 3x^5+15x^4+21x^3-15x^2-132x-180}{ x-2} = 3x^4+21x^3+63x^2+111x+90 $$Step 3:
The next rational root is $ x = -2 $
$$ \frac{ 3x^4+21x^3+63x^2+111x+90}{ x+2} = 3x^3+15x^2+33x+45 $$Step 4:
The next rational root is $ x = -3 $
$$ \frac{ 3x^3+15x^2+33x+45}{ x+3} = 3x^2+6x+15 $$Step 5:
The solutions of $ 3x^2+6x+15 = 0 $ are: $ x = -1+2i ~ \text{and} ~ x = -1-2i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.