Find roots of polynomial $$ p(x) = x^4+7x^3+\frac{61}{4}x^2+\frac{31}{2}x+10 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -2\\[1 em]x_2 &= -4\\[1 em]x_3 &= -\frac{ 1 }{ 2 }+i\\[1 em]x_4 &= -\frac{ 1 }{ 2 }-i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 4 } $.
$$ \begin{aligned} x^4+7x^3+\frac{61}{4}x^2+\frac{31}{2}x+10 & = 0 ~~~ / \cdot \color{blue}{ 4 } \\[1 em] 4x^4+28x^3+61x^2+62x+40 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -2 } $ is a root of polynomial $ 4x^4+28x^3+61x^2+62x+40 $.
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}{ 40 } $, with factors of 1, 2, 4, 5, 8, 10, 20 and 40.
The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 40 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 8, 10, 20, 40 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 40}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 40}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 40}{ 4} ~~ \end{aligned} $$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 }$
$$ \frac{ 4x^4+28x^3+61x^2+62x+40}{ x+2} = 4x^3+20x^2+21x+20 $$Step 3:
The next rational root is $ x = -2 $
$$ \frac{ 4x^4+28x^3+61x^2+62x+40}{ x+2} = 4x^3+20x^2+21x+20 $$Step 4:
The next rational root is $ x = -4 $
$$ \frac{ 4x^3+20x^2+21x+20}{ x+4} = 4x^2+4x+5 $$Step 5:
The solutions of $ 4x^2+4x+5 = 0 $ are: $ x = -\dfrac{ 1 }{ 2 }+i ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 }-i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.