1. ## R[x] - Reducible?

R[x] - Ring of polynomials in 'x' over Real numbers.
I read in my book that only elements of R[x] which might not be reducible further are that of degree 1,2.
It says proof is easy - and can be derived from the formula to solve quadratic equations.

I am lost - unable to derive the result? Any pointers please.

2. Here is an idea to get you started.

What about this polynomial in $\displaystyle \mathbb{R}[x]$
$\displaystyle x^2+1$.

See if this gets you started.

3. Originally Posted by TheEmptySet
Here is an idea to get you started.

What about this polynomial in $\displaystyle \mathbb{R}[x]$
$\displaystyle x^2+1$.

See if this gets you started.
Thanks for your post. It is -4, which doesn't have any square-root in real and hence $\displaystyle x^2+1$ has no real roots.

I guess I need lot more push ! I'm just not getting the idea of how to relate this to the proof above

4. Let's suppose that every element of $\displaystyle R[x]$ is reducible.

That means whenever $\displaystyle r \in R[x]$ and $\displaystyle r=ab, a,b \in R[x]$ and $\displaystyle a,b$ are not units.

That would mean that every degree 2 polynomial could be factored into two linear polynomial's with coeffeints in the reals.
i.e
$\displaystyle ax^2+bx+c=p(x+m)(x+n)$

But we know from the quadratic formula that is

$\displaystyle b^2-4ac < 0$ that the polynomial can only be factored if we use coeffeints from the complex numbers.

So if $\displaystyle b^2-4ac < 0$ then the only factorization over the real numbers would be of the form.

$\displaystyle ax^2+bx+c=n\left(\frac{a}{n}x^2+\frac{b}{n}x+\frac {c}{n}\right), n \ne 0$

Since $\displaystyle n \ne 0$ n is a unit in the reals(every non zero real number is a unit)

We can only factor it as a unit multiplied by some element of $\displaystyle R[x]$

5. Originally Posted by TheEmptySet
Let's suppose that every element of $\displaystyle R[x]$ is reducible.

That means whenever $\displaystyle r \in R[x]$ and $\displaystyle r=ab, a,b \in R[x]$ and $\displaystyle a,b$ are not units.

That would mean that every degree 2 polynomial could be factored into two linear polynomial's with coeffeints in the reals.
i.e
$\displaystyle ax^2+bx+c=p(x+m)(x+n)$

But we know from the quadratic formula that is

$\displaystyle b^2-4ac < 0$ that the polynomial can only be factored if we use coeffeints from the complex numbers.

So if $\displaystyle b^2-4ac < 0$ then the only factorization over the real numbers would be of the form.

$\displaystyle ax^2+bx+c=n\left(\frac{a}{n}x^2+\frac{b}{n}x+\frac {c}{n}\right), n \ne 0$

Since $\displaystyle n \ne 0$ n is a unit in the reals(every non zero real number is a unit)

We can only factor it as a unit multiplied by some element of $\displaystyle R[x]$
I follow this. What I am not able to understand/prove is that any polynomial in R[x] can be expressed as product of quadratic/linear polynomials.

For e.g. why a polynomial of degree 17 (or anything >2) can be factored into polynomials of degree 1 and 2?

Do we have to use the theorem of roots. That every polynomial has n-roots in complex, and complex roots occur in conjugates. then I can show the result. But I have nowhere used the quadratic formula as recommended in the book. I am wondering if I'm missing some imporant link?

6. Originally Posted by aman_cc
I follow this. What I am not able to understand/prove is that any polynomial in R[x] can be expressed as product of quadratic/linear polynomials.

For e.g. why a polynomial of degree 17 (or anything >2) can be factored into polynomials of degree 1 and 2?

Do we have to use the theorem of roots. That every polynomial has n-roots in complex, and complex roots occur in conjugates. then I can show the result. But I have nowhere used the quadratic formula as recommended in the book. I am wondering if I'm missing some imporant link?

Check you understand the following facts:

1) every real polynomial of odd degree has one real root

2) Complex non-real roots of real pol's appear in conjugate pairs, i.e.: if z = x + iy is a root of the real pol. f(x), then also z' = x - iy is a root of f(x) (basic knowledge of basic properties of complex conjugates is needed here)

Thus let f(x) in IR[x] of degree n >= 4 ==> we can write down its roots as follows: r_1, ...,r_n = real roots and z_1, z_1', z_2, z_2',...z_k, z_k' = complex non-real roots.
From here he can write
f(x) = (x-r_1)*...*(x-r_n)*(x-z_1)(x-z_1')*....

Now, check that every pair (x - z_i)(x - z_i') is a real root, and you get what you want.

Tonio

7. Originally Posted by tonio
Check you understand the following facts:

1) every real polynomial of odd degree has one real root

2) Complex non-real roots of real pol's appear in conjugate pairs, i.e.: if z = x + iy is a root of the real pol. f(x), then also z' = x - iy is a root of f(x) (basic knowledge of basic properties of complex conjugates is needed here)

Thus let f(x) in IR[x] of degree n >= 4 ==> we can write down its roots as follows: r_1, ...,r_n = real roots and z_1, z_1', z_2, z_2',...z_k, z_k' = complex non-real roots.
From here he can write
f(x) = (x-r_1)*...*(x-r_n)*(x-z_1)(x-z_1')*....

Now, check that every pair (x - z_i)(x - z_i') is a real root, and you get what you want.

Tonio
Thanks Tonio.