Find roots of polynomial $$ p(x) = 32x^{10}-33x^5+1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= \frac{ 1 }{ 2 }\\[1 em]x_3 &= 0.1545+0.4755i\\[1 em]x_4 &= 0.1545-0.4755i\\[1 em]x_5 &= -0.4045+0.2939i\\[1 em]x_6 &= -0.4045-0.2939i\\[1 em]x_7 &= -0.809+0.5878i\\[1 em]x_8 &= -0.809-0.5878i\\[1 em]x_9 &= 0.309+0.9511i\\[1 em]x_{10} &= 0.309-0.9511i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 32x^{10}-33x^5+1 $.
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}{ 1 } $, with factors of 1.
The leading coefficient is $ \color{red}{ 32 }$, with factors of 1, 2, 4, 8, 16 and 32.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 32 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 2, 4, 8, 16, 32 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} ~~ \pm \frac{ 1}{ 2} ~~ \pm \frac{ 1}{ 4} ~~ \pm \frac{ 1}{ 8} ~~ \pm \frac{ 1}{ 16} ~~ \pm \frac{ 1}{ 32} ~~ \end{aligned} $$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 }$
$$ \frac{ 32x^{10}-33x^5+1}{ x-1} = 32x^9+32x^8+32x^7+32x^6+32x^5-x^4-x^3-x^2-x-1 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ 32x^{10}-33x^5+1}{ x-1} = 32x^9+32x^8+32x^7+32x^6+32x^5-x^4-x^3-x^2-x-1 $$Step 3:
The next rational root is $ x = \dfrac{ 1 }{ 2 } $
$$ \frac{ 32x^9+32x^8+32x^7+32x^6+32x^5-x^4-x^3-x^2-x-1}{ 2x-1} = 16x^8+24x^7+28x^6+30x^5+31x^4+15x^3+7x^2+3x+1 $$Step 4:
Polynomial $ 16x^8+24x^7+28x^6+30x^5+31x^4+15x^3+7x^2+3x+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.