Find roots of polynomial $$ p(x) = 20x^4+9x^3+181x^2+81x+9 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 1 }{ 4 }\\[1 em]x_2 &= -\frac{ 1 }{ 5 }\\[1 em]x_3 &= 3i\\[1 em]x_4 &= -3i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 1 }{ 4 } } $ is a root of polynomial $ 20x^4+9x^3+181x^2+81x+9 $.
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}{ 9 } $, with factors of 1, 3 and 9.
The leading coefficient is $ \color{red}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 9 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1, 3, 9 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 9}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 9}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 9}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 9}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 3}{ 20} \pm \frac{ 9}{ 20} ~~ \end{aligned} $$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 }{ 4 } \right) = 0 $ so $ x = -\dfrac{ 1 }{ 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}{ 4x+1 }$
$$ \frac{ 20x^4+9x^3+181x^2+81x+9}{ 4x+1} = 5x^3+x^2+45x+9 $$Step 2:
The next rational root is $ x = -\dfrac{ 1 }{ 4 } $
$$ \frac{ 20x^4+9x^3+181x^2+81x+9}{ 4x+1} = 5x^3+x^2+45x+9 $$Step 3:
The next rational root is $ x = -\dfrac{ 1 }{ 5 } $
$$ \frac{ 5x^3+x^2+45x+9}{ 5x+1} = x^2+9 $$Step 4:
The solutions of $ x^2+9 = 0 $ are: $ x = 3 i ~ \text{and} ~ x = -3 i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.