Find roots of polynomial $$ p(x) = 20x^3+9x^2+13x+3 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 1 }{ 4 }\\[1 em]x_2 &= -\frac{ 1 }{ 10 }+\frac{\sqrt{ 59 }}{ 10 }i\\[1 em]x_3 &= -\frac{ 1 }{ 10 }- \frac{\sqrt{ 59 }}{ 10 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 1 }{ 4 } } $ is a root of polynomial $ 20x^3+9x^2+13x+3 $.
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}{ 3 } $, with factors of 1 and 3.
The leading coefficient is $ \color{red}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 3 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1, 3 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 3}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 3}{ 20} ~~ \end{aligned} $$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 }$
$$ \frac{ 20x^3+9x^2+13x+3}{ 4x+1} = 5x^2+x+3 $$Step 2:
The next rational root is $ x = -\dfrac{ 1 }{ 4 } $
$$ \frac{ 20x^3+9x^2+13x+3}{ 4x+1} = 5x^2+x+3 $$Step 3:
The solutions of $ 5x^2+x+3 = 0 $ are: $ x = -\dfrac{ 1 }{ 10 }+\dfrac{\sqrt{ 59 }}{ 10 }i ~ \text{and} ~ x = -\dfrac{ 1 }{ 10 }-\dfrac{\sqrt{ 59 }}{ 10 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.