Here is an idea to get you started.
What about this polynomial in
.
Now think about its disrciminant. i.e the part under the radical in the quadratic formula.
See if this gets you started.
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.
Let's suppose that every element of is reducible.
That means whenever and and 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
But we know from the quadratic formula that is
that the polynomial can only be factored if we use coeffeints from the complex numbers.
So if then the only factorization over the real numbers would be of the form.
Since 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
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