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Math Help - Irreducible polynomials degree 2 general form

  1. #1
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    Irreducible polynomials degree 2 general form

    Hi all!
    This should be a quick question.

    if f is an irreducible polynomial (on the real line) of degree two, i need to show that it can be written in the form f(x)= (x - a)^2 + b^2
    where a, and b lie on the real line and b is non-zero.

    ----------------------------------------------------------------------------------------------------------------------------------

    so, I know that irreducible polynomials have no real solutions...
    do I just assume f(x) can be written as x^2 + p*x+ q then complete the square and fudge it onto the expression above?
    or do i need to use eucledian division?

    many thanks
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  2. #2
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    Re: Irreducible polynomials degree 2 general form

    Yes, completing the square would give you (x+ p/2)^2+ q- p^2/4. Now consider three cases:
    q- p^2/4= 0
    q- p^2/4< 0 and
    q- p^2/4> 0

    Show that in the first and second cases, the polynomial can be factored with real coefficients but in the third, it cannot.
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  3. #3
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    Re: Irreducible polynomials degree 2 general form

    thanks
    so then would i say a=p/2 and b^2 = q - p^2/4 ??

    i also had to show the converse, like you suggested (another part of the qn). there I just expanded (x - a)^2 + b^2, equated to zero and used the quadratic formula, namely the discriminant, to show that f(x)=0 has no real roots.
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  4. #4
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    Re: Irreducible polynomials degree 2 general form

    i mean a = -p/2
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  5. #5
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    Re: Irreducible polynomials degree 2 general form

    first of all ANY monic real polynomial of degree 2 can be written in the form:

    (x - a)2 + c

    (this is what "completing the square" does).

    if this has a real root, then for some real x we have:

    (x - a)2 = -c

    now if c < 0, then we get:

    x - a = √(-c)

    x = a √(-c), as our 2 (and hence all) roots of (x - a)2 + c.

    if c = 0, we get a "double root" a (our polynomial is (x - a)2).

    however, if c > 0, then there is no real number x - a such that (x - a)2 = -c < 0.

    so if (x - a)2 + c is irreducible (which for a polynomial of degree 2 means no roots), c must be positive, in which case c = b2 for some real b (namely √c), that is:

    our polynomial is of the form (x - a)2 + b2.

    now the above only applies to MONIC polynomials, but any non-monic real polynomial is just a constant times a monic polynomial.

    for example:

    ax2 + bx + c = a(x2 + (b/a)x + c/a), and x2 + (b/a)x + c/a is monic.

    so your original statement isn't "quite true", for example, the polynomial:

    4x2 + 4 is irreducible over R (taking 4 out as a factor "doesn't count" because 4 is a UNIT (has a multiplicative inverse) in the field R), but cannot be put in the form you give.
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