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What Exactly are the Roots of a Polynomial?

  1. Jun 11, 2012 #1
    Are the roots of a polynomial given by the function f(x) defined as the values for x where f(x)=0?

    Does that mean f(x)=x^2 has only one root? Even though for every other value of x except zero there are two values for x that you can input to output a particular value for f(x).

    What about complex numbers?
     
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  3. Jun 11, 2012 #2

    micromass

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    You touch some very interesting points. Let's start easy. Take a polynomial f, we define a root of f as a point x such that f(x)=0.

    So, for example, [itex]f(x)=x^2[/itex] has one root, namely 0. But [itex]f(x)=x^2-1[/itex] has two roots, namely 1 and -1. And [itex]f(x)=x^2+1[/itex] has no real roots.

    As you noticed, this situation is unsatisfactory. Intuitively, we want [itex]x^2[/itex] to have two roots, not one. We want [itex]x^2[/itex] to have two equal roots. We say that the root 0 has multiplicity two.

    How is that defined? Well, take an arbitrary number a. We say that a is a root of the polynomial f of multiplicity n if we can write

    [tex]f(x)=(x-a)^ng(x)[/tex]

    for some polynomial g.

    Let's do some examples. Take a=0 and let [itex]f(x)=x+1[/itex]. Then we can only write

    [tex]f(x)=(x-0)^0(x+1)[/tex]

    so 0 is a root of multiplicity 0. We say that 0 is not a root.

    Take a=0 and let [itex]f(x)=x^2+x[/itex], then we can write

    [tex]f(x)=(x-0)^1(x+1)[/tex]

    so 0 is now a root of multiplicity 1. If we have [itex]f(x)=x^{44}(x+1)[/itex], then 0 is a root of multiplicity 44.

    We can see now that [itex]f(x)=x^2-1[/itex] has roots 1 and -1 and both have multiplicity 1. We can also see that [itex]f(x)=x^2[/itex] has only root 0 with multiplicity 2. And [itex]f(x)=x^2+1[/itex] still has no roots.

    Now we can easily prove that every polynomial of degree n has at most n roots, if we count with multiplicity. So a quadratic equation [itex]f(x)=x^2+bx+c[/itex] has at most 2 roots, counter with multiplicity. Of course, it can still happen that there are no roots.

    Another example: [itex]x^{100}+50x^4+3[/itex] has at most 100 roots, counter with multiplicity. So for example: there might by only 1 root with multiplicity 50, there might be 2 roots with multiplicity 50, there might be 50 roots with multiplicity 2, there might be 100 roots with multiplicity 1, etc.

    This was the real case. Now we can extend everything to complex numbers. Now we got something very beautiful. If we allow complex numbers to be roots, then a very famous theorem says: every polynomial of degree n has exactly n roots, counter with multiplicity. This is the "fundamental theorem of the algebra."

    For example, [itex]f(x)=x^2+1[/itex] has no real roots, but has two complex roots, namely i and -i. This is the nicest possible result.
     
  4. Jun 11, 2012 #3
    Yes, x2 has 1 root, which is said to be of "multiplicity 2" in order to preserve the fact that a second degree poly should have two roots.

    Every (nonconstant) polynomial of degree n with complex coefficients has exactly n roots in the complex numbers ... as long as you count multiplicities to account for roots that appear more than once in the linear factorization of the polynomial.

    This fact is known as the Fundamental Theorem of Algebra ...

    http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

    The first widely accepted proof was published by Gauss in 1799 as his Ph.D. thesis when he was 21 years old.

    It's a curiosity that the FTA is really a theorem of analysis, not algebra; as its proof depends on the topological completeness of the complex plane.
     
  5. Jun 11, 2012 #4
    The roots of a polynomial f(x) are the values for which f(x) = 0, yes.
    I'm not sure what you mean by "what about complex numbers"; you can define polynomials over the field of complex numbers as well, if you like. It's a rather important theorem that any polynomial with coefficients in the complex numbers (which includes the reals) has a root in the complex numbers. Note that this is not necessarily true of the reals; the polynomial x^2 + 1, for instance, does not have any real roots.
     
  6. Jun 11, 2012 #5
    Yes, they're values of x so that the whole expression is equal to zero.

    For example, the polynomial x2 + 4 has no real roots, but it has 2i as a root.
     
  7. Jun 11, 2012 #6

    Mentallic

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    Yes.
    Yes, basically. But there is a particularly important theorem that states that for any polynomial of nth degree, there are exactly n complex roots (but not all necessarily distinct [different]).
    The underlined part means that you can have a polynomial of nth degree, but it only has (n-1) roots. This means there must be a double root somewhere, just as you've noticed with the function f(x)=x2.

    It's very common for polynomials to have 2 or more x-values that output to give the same f(x) value. Do you know about inverse functions?

    Look back at the theorem and notice the part about complex roots. Every real number is a subset of a complex number (so it can be said that a real number is also complex), it's just that in its complex form z=a+ib, the value of b=0.
    If we have the function f(x)=x2+1, while there don't exist any real values for x such that f(x)=0, there certainly exist complex numbers. Mainly [itex]x=\pm i[/itex].
    And again notice that the degree of f(x) is 2, and as it so happens, we have 2 complex roots which in this case are distinct.


    EDIT: haha so many replies slipped in before me :biggrin:
     
  8. Jun 12, 2012 #7
    Graphically, real roots of polynomial is where the graph cuts the x line. :)
     
  9. Jun 12, 2012 #8

    HallsofIvy

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    Strictly speaking you should not use the phrase "roots of a polynomial". An equation has "roots", a polynomial has "zeros". The zeros of the polynomial p(x) are the roots of the equation p(x)= 0.

    Unfortunately, people are seldom that strict!
     
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