Find roots of polynomial $$ p(x) = 34+81x+60x^2+14x^3+2x^4+x^5 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -1\\[1 em]x_2 &= -2\\[1 em]x_3 &= -1\\[1 em]x_4 &= 1+4i\\[1 em]x_5 &= 1-4i \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} 34+81x+60x^2+14x^3+2x^4+x^5 & = 0\\[1 em] x^5+2x^4+14x^3+60x^2+81x+34 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -1 } $ is a root of polynomial $ x^5+2x^4+14x^3+60x^2+81x+34 $.
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}{ 34 } $, with a single factor of 1, 2, 17 and 34.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 34 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 17, 34 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 17}{ 1} \pm \frac{ 34}{ 1} ~~ \end{aligned} $$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( -1 \right) = 0 $ so $ x = -1 $ 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+1 }$
$$ \frac{ x^5+2x^4+14x^3+60x^2+81x+34}{ x+1} = x^4+x^3+13x^2+47x+34 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ x^5+2x^4+14x^3+60x^2+81x+34}{ x+1} = x^4+x^3+13x^2+47x+34 $$Step 4:
The next rational root is $ x = -2 $
$$ \frac{ x^4+x^3+13x^2+47x+34}{ x+2} = x^3-x^2+15x+17 $$Step 5:
The next rational root is $ x = -1 $
$$ \frac{ x^3-x^2+15x+17}{ x+1} = x^2-2x+17 $$Step 6:
The solutions of $ x^2-2x+17 = 0 $ are: $ x = 1+4i ~ \text{and} ~ x = 1-4i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.