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Thread: R[x] - Reducible?

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    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.
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    Here is an idea to get you started.

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

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

    See if this gets you started.
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    Quote Originally Posted by TheEmptySet View Post
    Here is an idea to get you started.

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

    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 $\displaystyle x^2+1$ 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
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    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 $\displaystyle R[x]$
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    Quote Originally Posted by TheEmptySet View Post
    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?
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    Quote Originally Posted by aman_cc View Post
    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
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    Quote Originally Posted by tonio View Post
    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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