Polynomials which factors but have no roots?

**1337h4x** · May 28th 2010, 04:53 AM

Hello all,

Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

Any help is appreciated!

**chiph588@** · May 28th 2010, 07:36 AM

Originally Posted by 1337h4x

Hello all,

Can anyone think of an example of a polynomial f(g) with integer coefficients which factors (poly mod n) but has no roots, e.g. for which there are no such integers g where f(g)=0(mod n) ?

What are appropriate values of n, the coefficients of f(g), and how can I show that it has no roots?

Any help is appreciated!

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

Now consider this polynomial modulo $2$ .

**1337h4x** · May 28th 2010, 07:55 AM

Originally Posted by chiph588@

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

Now consider this polynomial modulo $2$ .

Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

I'm a little confused about this as you can tell...

**chiph588@** · May 28th 2010, 08:07 AM

Originally Posted by 1337h4x

Thank you ! How can I prove that this has no roots though? I can't just say it can't be factored any further with integers, or can I? Do I just find the quadratic roots and show that they aren't integers? Aren't those numbers still roots? Or how does this have no roots still?

I'm a little confused about this as you can tell...

Just plug in $0$ and $1$ into the polynomial.

**1337h4x** · May 28th 2010, 08:15 AM

Originally Posted by chiph588@

Just plug in $0$ and $1$ into the polynomial.

If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?

**chiph588@** · May 28th 2010, 08:23 AM

Originally Posted by 1337h4x

If you plug in 0 you get 2, if you plug in 1 you get three. What is your point?

No when you plug in $0$ you get $1$ .

Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.

**wonderboy1953** · May 28th 2010, 08:27 AM

All polynomials have roots. The question is determining those roots.

**chiph588@** · May 28th 2010, 08:31 AM

Originally Posted by wonderboy1953

All polynomials have roots. The question is determining those roots.

Over the complex numbers yes, but not an arbitrary field like $\mathbb{Z}/2\mathbb{Z}$ .

**1337h4x** · May 28th 2010, 08:43 AM

Originally Posted by chiph588@

No when you plug in $0$ you get $1$ .

Anyway those are all the elements modulo 2 and none of them gave you a root, but you can still factor your polynomial.

Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?

**chiph588@** · May 28th 2010, 09:01 AM

Originally Posted by 1337h4x

Okay, so when I plug in 0 I get 1, when I plug in 1 I get 3, so how are the results of 1 and 3 relate to modulo 2 ? Is it because 2|(3-1) ? Is that what you mean?

And from what you and wonderboy are discussing, I take it that when we say there are no roots, we mean that there are no REAL roots? no integer roots? Both?

Modulo 2 there are no roots what so ever.

Let $f(x)=x^4+x^2+1$ .

$f(0)=1\not\equiv0\bmod{2}$
$f(1)=3\equiv1\not\equiv0\bmod{2}$

Thus there are no roots modulo $2$ .

**1337h4x** · May 28th 2010, 09:17 AM

Originally Posted by chiph588@

Modulo 2 there are no roots what so ever.

Let $f(x)=x^4+x^2+1$ .

$f(0)=1\not\equiv0\bmod{2}$
$f(1)=3\equiv1\not\equiv0\bmod{2}$

Thus there are no roots modulo $2$ .

So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?

**chiph588@** · May 28th 2010, 09:27 AM

Originally Posted by 1337h4x

So how do you consider the system "modulo 2" then? I was looking back at your work, and then you stated that it was modulo 2. How ?

Read your original question. I showed you a polynomial f(x) that can be factored but there is no g such that $f(g)\equiv0 \bmod{2}$ .

**1337h4x** · May 28th 2010, 09:37 AM

Ah, I see what you're saying. Thanks for making the connection. I lost track of what I was originally asking apparently!

Thread: Polynomials which factors but have no roots?

LinkBack

Thread Tools

Display

Polynomials which factors but have no roots?

Answer

Similar Math Help Forum Discussions

Roots of 3rd degree polynomials and Characteristic Polynomials

prove a simple property of factors of polynomials

[SOLVED] Complex factors of polynomials

A polynomial that factors (poly mod n) but has no roots?

an example of a ploynomial which factors, but has no roots

Search Tags