Proof of factor theorem

Discussion in 'Physics & Math' started by Fudge Muffin, Nov 3, 2012.

  1. Fudge Muffin Fudge Muffin Registered Senior Member

    Messages:
    148
    *if someone could link me to a guide on how to use those really really cool alegra notation thingys it would be a great help

    but yeah, the proof follows on from the proof of the remainder theorem: The remainer when (X-a) is divided into f(X) is f(a)

    How to prove this... I was reading a book that said let f(X) and g(X) be two polynomials, where deg f(X) is greater than or equal to deg g(X), and let q(X) be:

    q(X) = f(X)/g(X)

    and so f(X) = q(X)g(X) + r(X) where r(X) is the remainder term if it doesn't divide exactly... ok...

    then suppose deg g(X) = 1 then we can write g(X) as (X-a) where a is a constant, and the remainder will have a degree of less than 1 (what does this mean? is the degree zero? but anyway it's a constant so let it = c) and we have

    f(X) = q(X)(X-a) + c

    then it says evaluate both sides, and bam, you've proved it. Um, yeah, how exactly do I evaluate both sides?

    Also, WHAT exactly is a 'zero-polynomial' ?
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    A general polynomial \(f(x) = \sum_{n=1}^{N} a_{n}x^{b_{n}}\) (for \(b_{n} \in \{0,1,\ldots\}\) non-negative integers) is degree D if \(\max(b_{n}) = D\). Therefore the degree 0 polynomial requires \(\max(b_{n}) = 0\) and since \(x^{0} = 1\) you have \(f(x) = a\) is the general degree 0 polynomial. The particular special case a=0, so f(x) = 0, is also a polynomial and you might even call it 'the zero polynomial' in the same way you might call \(\mathbf{v} = (0,0,\ldots,0)\) 'the zero vector' in a \(\mathbb{R}^{N}\) vector space.

    As for 'evaluating each side' you have what is supposed to be an identity, true for all values of X. You don't know anything about q(X) and you need to work out c, which is a constant. Well you know that X-a=0 when X=a so if you evaluate both sides at X=a then you will have the left hand side f(a) and the right hand side as q(a)(a-a) + c = c. So now you have f(a) = c. So now you have f(X) = q(X)(X-a) + f(a), since it was an identify. There's the result you wanted

    Please Register or Log in to view the hidden image!

     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Fudge Muffin Fudge Muffin Registered Senior Member

    Messages:
    148
    awh brilliant! thank you! been plaguing my mind for like 2 months, that's such a nice step! aaaawh!

    Also, i read the 'degree' of a non-zero polynomial is minus infinity. the book tried to explain why but it didn't make sense, it actually said 'it's best not to think of this too much' i assume it would just be for convenience in certain theorems etc.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.

Share This Page