Find all roots of these two polynomials

84 Views Asked by At

There are only answers without any reasons in my textbook's questions.

PLEASE HELP ME :(

Find the root of these polynomials by using hints.

Let the $w$ is complex root of $x^2+x+1$

1.First question

$f(x)=x^3 + 6x^2 -2 $ which means $ irr(\alpha,Q)$ (hint : one of the real root of $f(x)$ is $\alpha = \sqrt[3]{4}-\sqrt[3]{2}$ )

Solution in my textbook said that that all the roots are

$\alpha$

$\beta$ =$\sqrt[3]{4}w^2-\sqrt[3]{2}w$,

$\gamma$ = $\sqrt[3]{4}w-\sqrt[3]{2}w^2$'

I'm tried to find the other roots $\beta$ and $\gamma$ But failed. :(

Is there any Either formula or principle finding the $\beta$ and $\gamma$? Please give me some ideas.

2.Second question

$g(x)=x^9 -3x^6+165x^3-1 $ which means $ irr(\alpha,Q)$ (hint : one of the real root of $g(x)$ is $\alpha = \sqrt[3]{3}-\sqrt[3]{2}$ )

Solution in my textbook said that that all the roots form is

$ \sqrt[3]{3}w^n-\sqrt[3]{2}w^m$ [ $m,n=0,1,2$]

All I know the number of roots is 9 thinking the splitting field of $g(x)$ over $Q$ And It might having the similar principle with the case of $f(x)$

BUT I couldn't find exact all the complex roots though $\alpha$ was given.

Why is all the roots of $g(x)$ have a form that $ \sqrt[3]{3}w^n-\sqrt[3]{2}w^m$?

2

There are 2 best solutions below

3
On BEST ANSWER

Let $\theta = \sqrt[3]{2}$. Then the conjugates of $\theta$ are $\theta\omega$ and $\theta\omega^2$, where $\omega$ is a primitive cubic root of unity.

Now $\alpha = \theta^2-\theta = h(\theta)$ and so its conjugates are $h(\theta\omega)$ and $h(\theta\omega^2)$. These are the other roots of $f$.

A similar answer works for the second question.

2
On

Divide $$(x^3+6x^2-2):(x-(\sqrt[3]{4}-\sqrt[3]{2}))=$$ and you will get a quadratic one.