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Complex Polynomial Theorem

  1. May 5, 2004 #1
    Can someone help me prove this?

    Where f(z) is a complex polynomial of degree n, and g(z) is a complex polynomial of degree n-1, and f(q)=0, I can't seem to prove


    I have substitituted for each of the polynomials and distributed out the z and q terms, but I am unsure how to manipulate this equation. Help anyone?

    I am at the point so far where I have

    anz^n + an-1z^n-1 +...+a1z+a0= bn-1z^n +bn-2z^n-1+...b1z^2+b0z-q(bn-1z^n-1+bn-2z^n-2+...b0)

    But unable to manipulate this further.
  2. jcsd
  3. May 5, 2004 #2
    I think this is a standard theorem (or a variation of one) that you can find in any textbook on abstract algebra or polynomials. It isn't really a complex analysis problem.
  4. May 5, 2004 #3


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    The first thing you should do is state the theorem clearly. The way you put it, it could be interpreted as saying that "If f(z) is any nth order polynomial such that f(q)= 0 and g(z) is any n-1 order polynomial then f(z)= (z-q)g(z)" and that, of course, is not true.

    The theorem you want says that "If f(z) is any nth order polynomial such that f(q)= 0 then THERE EXIST an n-1 order polynomial g such that f(z)= (z-a)g(z)". To do that, first use the "Euclidean Algorithm" to prove that, for any function f(z), over the complex numbers and any complex number q, There exist a function g(z) and a constant a such that f(z)= (z-a)q(z)+ a.
  5. May 5, 2004 #4
    Thanks for your reply. I am just not sure how to apply the Euclidan algorithm to something other than numbers. Since f(q) or f(a) as you call it equals zero, can we say that a divides f(z)?
  6. May 6, 2004 #5


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    The Euclidean algorithm, in it's simplest form, says that if a and b are any two numbers, then there exist two number Q and r such that
    b= aQ+ r where (absolute value of) r is less than (absolute value of) a.

    That is essentially "division". Q is the quotient of b divided by a and r is the remainder.

    However, it is also true for polynomials: If a and b are any polynomials, then there exist polynomials Q and r such that b= aQ+ r and the degree of r is less than the degree of a.

    In particular, if a= x-q so that it has degree 1, then r must have degree 0 and so is a constant (a number).

    Also, in this example, b= f(z), a= (z-q) so f(z)= (z-q)Q(z)+ r and f(q)= 0*Q(z)+ r= 0 so r must equal 0.
  7. Jul 23, 2004 #6


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    I am delighted that you are studying this question. This is an old high school algebra theorem formerly called the "root - factor" theorem. I.e. if f(x) is a polynomial (could be with real or rational or even integral coefficients) then q is a root of f (i.e. f(q) = 0), if and only if x-q is a factor of f(x).

    This is very basic to all algebraic manipulations used throughout mathematics, but I have essentially never had a single student in 25 years of teaching college who knew this principle, although it was taught in all high school algebra courses in the 1950's.

    As Halls of Ivy explained it follows just by dividing f(x) by x-q and noticing that you can keep dividing until the remainder is a constant r. Then you have f(x) = (x-q)g(x) +r, for some polynomial g(x), whose degree must be one less than deg(f). Then if you plug in x=q you will see that f(q) = r. Hence f(q) = 0 if and only if the remainder r after division by x-q was zero, if and only if x-q divided f(x) evenly.

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