1. ## counter examples!

I just need a counter example for each of theses problems, they have been killing me for a few days now..

A) The gcd of

$2x^2+4x+2$
$4x^2 +12x+8$

in $\mathbb{Q}[x]$ is $2x+2$

B) If $k$ is a field in $p(x) \in k[x]$ is a non constant polynomial having no roots in $k$, then $p(x)$ is irreducible in $k[x]$

These are both false, but Everything i try works for them...

2. Originally Posted by ux0
I just need a counter example for each of theses problems, they have been killing me for a few days now..

A) The gcd of

$2x^2+4x+2$
$4x^2 +12x+8$

in $\mathbb{Q}[x]$ is $2x+2$

$2x^2+4x+2=2(x+1)^2\,,\,4x^2+12x+8=4(x+1)(x+2)$, so the gcd is $x+1$ , up to a constant, and thus your answer is correct, though perhaps not the prettiest one.

B) If $k$ is a field in $p(x) \in k[x]$ is a non constant polynomial having no roots in $k$, then $p(x)$ is irreducible in $k[x]$

$(x^2+1)^5$ has no roots over $\mathbb{R}$, but it is terribly NOT irreducible...

Tonio

These are both false, but Everything i try works for them...
.

3. THose solution don't work, each of those statements are Actually False, so I need to find an example that proves them wrong...

So in the first part, i need to find the GCD that's $> 2x+2$

And in the second part, i need to find a poly, that is in a Field, $k$. Which has no roots in $k$, but it is reducible in the $k[x]$

that's the problem I'm having...

I'm thinking in part two I need to use the proposition:

If k is a field, then every non constant polynomial $f(x) \in k[x]$ has a factorization

$f(x)=ap_1(x)....p_t(x)$

where a is a nonzero constant and the $p_i(x)$ are monic irreducible polynomials.
or this proposition

Let $k$ be a field and let $f(x) \in k[x]$ be a quadratic or cubic polynomial. Then $f(x)$ is irreducible in k[x] if and only if $f(x)$ does not have a root in $k$
It is clear by the second proposition my counter example to part two would be of degree greater than 3...

4. oops found part two...

$x^4+2x+1 = (x^2+1)^2$

No real roots, but is reducible..

Still don't have part 1... i was thinking it was like you said

2x+2 is just a linear combo of x+1 therefore x+1 is unique, and monic, where 2x+2 is not unique.

5. Originally Posted by ux0
oops found part two...

$x^4+2x+1 = (x^2+1)^2$

No real roots, but is reducible..

Still don't have part 1... i was thinking it was like you said

2x+2 is just a linear combo of x+1 therefore x+1 is unique, and monic, where 2x+2 is not unique.

I don't know what you're doing: in my previous post I answered your questions. You may believe what you want but unless proven wrong I'm certain 99.999% about their being right.

Tonio