# Thread: IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

1. ## IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

Attempt:

Ive shown that this has no linear factors as +-1 +-2 are not roots.

Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

And I end up with
a +c=1
d+ac+b=-1
bd=-2

But i can seam to show why no a b c d exist (or Exist)

Is there an easier approach?

I have an exam in 12 hours so I hope soeone can help!

Niall

2. Originally Posted by Niall101
bd=-2
It is sufficient to assume irreduciblility over Z by Gauss's Lemma.
Now this condition tells us that $\displaystyle b=\pm 1 \text{ and }d=\mp 2$ or $\displaystyle b=\pm 2 \text{ and }d=\mp 1$.
Show that any one of these leads to a contradiction i.e. those equations are not solvable with integers.

3. Great thanks! Its so obvious now! I ment to say a b c d were in Z. Thats what a week of no sleep will do to you

4. Originally Posted by Niall101
IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

Attempt:

Ive shown that this has no linear factors as +-1 +-2 are not roots.

Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

And I end up with
a +c=1
d+ac+b=-1
bd=-2

But i can seam to show why no a b c d exist (or Exist)

Is there an easier approach?

I have an exam in 12 hours so I hope soeone can help!

Niall
i will note this theorem.
If $\displaystyle f(x)$ is reducible over $\displaystyle \mathbb{Q}$, then $\displaystyle f(x)$ is reducible over $\displaystyle \mathbb{Z}$.

so, as a contrapositive of the statement,
If $\displaystyle f(x)$ is irreducible over $\displaystyle \mathbb{Z}$, then $\displaystyle f(x)$ is irreducible over $\displaystyle \mathbb{Q}$.

from this, $\displaystyle a,b,c,d \in \mathbb{Z}$

$\displaystyle bd = -2$
then WLOG, $\displaystyle b = \pm 1$ and $\displaystyle d=\mp 2$.

try to solve the combinations..

EDIT: aww, didn't notice that there were replies already..

5. with a=1 c=0 b=1 d=-2 I get (x^2+x+1)(x^2-2)=F(x) so its not irreducable! Thanks for all the replies! Hope I can do this in the exam later!

6. Sorry for another question. If i have F(x)=x^3-5 Is it sufficiant to show it have no linear factors by plugging in +-5 +-1 and showing not equal to zero? I was thinking because the only factors could be linear or of degree 3. Thanks again.

7. yes, it is enough..

8. Hey thanks so much! I really need to get this exam to pass the year. I really appreciate your help and such quick response

9. If anyone has a chance to shed some light on this one it would be a great help

Let A = cuberoot(5),
and let
F = a + bA + cA in R
a; b; c in Q
Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of A?

10. Originally Posted by Niall101
If anyone has a chance to shed some light on this one it would be a great help

Let A = cuberoot(5),
and let
F = a + bA + cA in R
a; b; c in Q
Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of A?
should that red thing be squared?
assuming yes,
$\displaystyle A = \sqrt[3]{5}$

$\displaystyle F = a + bA + cA^2 \subseteq \mathbb{R}$ where $\displaystyle a,b,c \in \mathbb{Q}$

what we want is to find $\displaystyle a + bA + cA^2$ so that $\displaystyle (2 +A +A^2)(a + bA + cA^2)=1$

if we try to multiply, we will get $\displaystyle 2a + (2b+a)A + (a+2c+b)A^2 + (c+b)A^3 + (c)A^4 = 1$

however, $\displaystyle A^4 = A^3A = 5A$

thus, we reduce it to $\displaystyle 2a + (2b+a + 5c)A + (a+2c+b)A^2 + (c+b)A^3 = 1$

can you do it from here?

11. Yes I think I have it now. I actually had this a.A^3=5.A written down on one of my attempts. Thank you so much for all the help. Your a life saver!

12. Hey guys thanks again! The exam went great! only 2 more to go

13. good for you!
if you still have questions, don't hesitate to post it.. (but on another thread.. )

14. Great thanks! On my way out now to face Group theory!