Find roots of polynomial $$ p(x) = 2x^{10}+7x^9-4x^8-2x^2-7x+4 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1\\[1 em]x_3 &= -4\\[1 em]x_4 &= \frac{ 1 }{ 2 }\\[1 em]x_5 &= 0.7071+0.7071i\\[1 em]x_6 &= 0.7071-0.7071i\\[1 em]x_7 &= -0.7071+0.7071i\\[1 em]x_8 &= -0.7071-0.7071i\\[1 em]x_9 &= i\\[1 em]x_{10} &= -i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 2x^{10}+7x^9-4x^8-2x^2-7x+4 $.

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}{ 4 } $, with factors of 1, 2 and 4.

The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 4 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 4 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} ~~ \end{aligned} $$

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 }$

$$ \frac{ 2x^{10}+7x^9-4x^8-2x^2-7x+4}{ x-1} = 2x^9+9x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+3x-4 $$

Step 2:

The next rational root is $ x = 1 $

$$ \frac{ 2x^{10}+7x^9-4x^8-2x^2-7x+4}{ x-1} = 2x^9+9x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+3x-4 $$

Step 3:

The next rational root is $ x = -1 $

$$ \frac{ 2x^9+9x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+3x-4}{ x+1} = 2x^8+7x^7-2x^6+7x^5-2x^4+7x^3-2x^2+7x-4 $$

Step 4:

The next rational root is $ x = -4 $

$$ \frac{ 2x^8+7x^7-2x^6+7x^5-2x^4+7x^3-2x^2+7x-4}{ x+4} = 2x^7-x^6+2x^5-x^4+2x^3-x^2+2x-1 $$

Step 5:

The next rational root is $ x = \dfrac{ 1 }{ 2 } $

$$ \frac{ 2x^7-x^6+2x^5-x^4+2x^3-x^2+2x-1}{ 2x-1} = x^6+x^4+x^2+1 $$

Step 6:

Polynomial $ x^6+x^4+x^2+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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