Factorize in Q[x] , R[x] and C[X]

[kezman]

Factorize in Q[x] , R[x] and C[X]

$$4x^6 + 8x^5 - 3x^4 - 19x^3 - 26x^2 - 15x - 3$$


I couldn't finish it but this is up to where I did:

$$4(x+1/2)^2 (x^4+x^3-2x^2-3x-3)$$

 

[arildno]

Your factorization cannot be correct.
In the original one, 1 is not a root, whereas 1 is a root in your second one.

 

[kezman]

Im sorry I can't see that.
In the second gives -6

 

[arildno]

Sorry, my mistake.

 

[0rthodontist]

Well, my calculator says that the remaining real solutions of your fourth-degree factor are $$\pm \sqrt{3}$$ but I don't know how you could find those by hand.

 

[kezman]

Yes that's right thanks

 

[kezman]

but I don't know how you could find those by hand.

That was a problem in my exam

 

[0rthodontist]

Well--how do you find them by hand then, seeing how they are irrational?

 

[StatusX]

It's easy to check if a polynomial has any rational roots. To see if it has irrational roots of the form $\sqrt{r}$ for r rational, you can break up the polynomial into even and odd terms and substitute in $\sqrt{r}$, which will leave something like $e(r)+o(r) \sqrt{r}$. For r rational, this will only be zero if e(r)=o(r)=0. In this case,

$$x^4+x^3-2x^2-3x-3 = (x^4+2x^2-3) + (x^3-3x)=0$$

plugging in x= $\sqrt{r}$:

$$(r^2+2r-3) + \sqrt{r} (r-3)$$

We see that r=3 is a solution to both terms, and so $x=\pm \sqrt{3}$ is a root of the original polynomial. This process can be generalized to cube roots, etc, but I doubt if it's very useful unless you know in advance that you'll have a root of this form.