Find roots of polynomial $$ p(x) = x^6+x^5-5x^4-2x^3-x^2-x+6 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= 0.9421\\[1 em]x_3 &= -1.1704\\[1 em]x_4 &= -2.6081\\[1 em]x_5 &= -0.0818+1.0181i\\[1 em]x_6 &= -0.0818-1.0181i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ x^6+x^5-5x^4-2x^3-x^2-x+6 $.
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}{ 6 } $, with a single factor of 1, 2, 3 and 6.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 6 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} ~~ \end{aligned} $$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( 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 }$
$$ \frac{ x^6+x^5-5x^4-2x^3-x^2-x+6}{ x-2} = x^5+3x^4+x^3-x-3 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ x^6+x^5-5x^4-2x^3-x^2-x+6}{ x-2} = x^5+3x^4+x^3-x-3 $$Step 3:
Polynomial $ x^5+3x^4+x^3-x-3 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.