Find roots of polynomial $$ p(x) = 2x^6+3x^3-5 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1.3572\\[1 em]x_3 &= 0.6786+1.1754i\\[1 em]x_4 &= 0.6786-1.1754i\\[1 em]x_5 &= -0.5+0.866i\\[1 em]x_6 &= -0.5-0.866i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 2x^6+3x^3-5 $.
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}{ 5 } $, with factors of 1 and 5.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} ~~ \end{aligned} $$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 }$
$$ \frac{ 2x^6+3x^3-5}{ x-1} = 2x^5+2x^4+2x^3+5x^2+5x+5 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 2x^6+3x^3-5}{ x-1} = 2x^5+2x^4+2x^3+5x^2+5x+5 $$Step 3:
Polynomial $ 2x^5+2x^4+2x^3+5x^2+5x+5 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.