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Lower and Upper bounds of Polynomial equations

  1. Feb 15, 2013 #1
    Recently I am studying about theorems regarding to polynomial equations and encounter the lower and upper bounds theorem. Which states that if a<0 and P(a) not equals 0, and dividing P(x) by (x-a) leads to coefficients that alternate signs, then a is a lower bound of all the roots of P(x)=0. The proof about this statement is provided but I have troubles in understanding it(I do understand about the proof of upper bound one...), I hope someone here can explain and proof about the theorem... Thanks for the help!!!
  2. jcsd
  3. Feb 15, 2013 #2
    It would help us enormously if you could show the proof (or a reference) and indicate which part you don't understand.
  4. Feb 15, 2013 #3
    (Lower Bound)(Let P(x) be any polynomial with real coefficients and a positive leading coefficient.)
    Let r be a root of the given polynomial. Assuming r is not equal to 0 and that the polynomial has a nonzero constant term. Since P(r)=0, we can say that P(x)=(x-r)Q(x). Substituting in x=a, we get that P(a)=(a-r)Q(a).
    Since P(a),Q(a) do not equal to 0, we can divide both sides by Q(a) to get (a-r) = P(a)/Q(a).
    We know that Q(a) is either positive or negative. Since a<0 and the leading terms in Q(x) has a positive coefficient, the constant term in Q(x) has the same sign as Q(a).
    If we assume that neither r nor the constant term in P(x) are zero, then that guarantees that the constant term in Q(x) must be strictly positive or negative. [Since the coefficients alternate in sign, P(a) differs in sign from the constant term in Q(x). But since the constant term in Q(x) has the same sign as Q(a) we know that P(a) and Q(a) are opposite signs, which implies that (a-r)<0, which leads to a<r.]
    Sentences in square brackets are the parts which I don't understand.
  5. Feb 15, 2013 #4
    Last edited by a moderator: May 6, 2017
  6. Feb 16, 2013 #5
    can someone help, please? :cry:
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