Find roots of polynomial $$ p(x) = 5x^6-19x^5-169x^4-283x^3-160x^2-22x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -1\\[1 em]x_3 &= -\frac{ 1 }{ 5 }\\[1 em]x_4 &= -1\\[1 em]x_5 &= 8.5678\\[1 em]x_6 &= -2.5678 \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ 5x^6-19x^5-169x^4-283x^3-160x^2-22x $ and solve two separate equations:
$$ \begin{aligned} 5x^6-19x^5-169x^4-283x^3-160x^2-22x & = 0\\[1 em] \color{blue}{ x }\cdot ( 5x^5-19x^4-169x^3-283x^2-160x-22 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 5x^5-19x^4-169x^3-283x^2-160x-22 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Use rational root test to find out that the $ \color{blue}{ x = -1 } $ is a root of polynomial $ 5x^5-19x^4-169x^3-283x^2-160x-22 $.
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}{ 22 } $, with factors of 1, 2, 11 and 22.
The leading coefficient is $ \color{red}{ 5 }$, with factors of 1 and 5.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 22 }}{\text{ factors of 5 }} = \pm \dfrac{\text{ ( 1, 2, 11, 22 ) }}{\text{ ( 1, 5 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 11}{ 1} \pm \frac{ 22}{ 1} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 11}{ 5} \pm \frac{ 22}{ 5} ~~ \end{aligned} $$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 }$
$$ \frac{ 5x^5-19x^4-169x^3-283x^2-160x-22}{ x+1} = 5x^4-24x^3-145x^2-138x-22 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ 5x^5-19x^4-169x^3-283x^2-160x-22}{ x+1} = 5x^4-24x^3-145x^2-138x-22 $$Step 4:
The next rational root is $ x = -\dfrac{ 1 }{ 5 } $
$$ \frac{ 5x^4-24x^3-145x^2-138x-22}{ 5x+1} = x^3-5x^2-28x-22 $$Step 5:
The next rational root is $ x = -1 $
$$ \frac{ x^3-5x^2-28x-22}{ x+1} = x^2-6x-22 $$Step 6:
The solutions of $ x^2-6x-22 = 0 $ are: $ x = 3-\sqrt{ 31 } ~ \text{and} ~ x = 3+\sqrt{ 31 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.