R[x] - Reducible?

**aman_cc** · October 18th 2009, 06:51 AM

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.

**TheEmptySet** · October 18th 2009, 07:19 AM

Here is an idea to get you started.

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

Now think about its disrciminant. i.e the part under the radical in the quadratic formula.

See if this gets you started.

**aman_cc** · October 18th 2009, 07:25 AM

Originally Posted by TheEmptySet

Here is an idea to get you started.

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

Now think about its disrciminant. i.e the part under the radical in the quadratic formula.

See if this gets you started.

Thanks for your post. It is -4, which doesn't have any square-root in real and hence $x^2+1$ 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

**TheEmptySet** · October 18th 2009, 07:51 AM

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

That means whenever $r \in R[x]$ and $r=ab, a,b \in R[x]$ and $a,b$ 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
$ax^2+bx+c=p(x+m)(x+n)$

But we know from the quadratic formula that is

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

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

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

Since $n \ne 0$ 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 $R[x]$

**aman_cc** · October 18th 2009, 07:58 AM

Originally Posted by TheEmptySet

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

That means whenever $r \in R[x]$ and $r=ab, a,b \in R[x]$ and $a,b$ 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
$ax^2+bx+c=p(x+m)(x+n)$

But we know from the quadratic formula that is

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

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

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

Since $n \ne 0$ 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 $R[x]$

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?

**tonio** · October 18th 2009, 10:17 AM

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

**aman_cc** · October 18th 2009, 08:27 PM

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.

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