1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polynomial factorization

  1. Nov 27, 2003 #1
    question :

    if p is a common factor of the equations f(x)=0, and g(x)=0, prove that p is also a root of the equation af(x) + bg(x) where a and b are constants.


    i want to know the working.
    i dont think the working will be like this :

    af(p) + bg(p) = a(0) + b(0) = 0

    so, i need the correct working here. thanks.
     
  2. jcsd
  3. Nov 27, 2003 #2
    What is a 'factor of an equation'? I don't understand...
     
  4. Nov 27, 2003 #3

    AD

    User Avatar

    Surely you must know what factorisation of polynomials is, Arcnets.
     
  5. Nov 27, 2003 #4
    Yeah, OK. So f(x) and g(x) share the factor (x-p), right?
    [tex]
    f(x) = 0 \ (\bmod{(x-p)}) \ \wedge \ g(x) = 0 \ (\bmod{(x-p)})
    [/tex]
    [tex]
    \Rightarrow
    [/tex]
    [tex]
    af(x) + bg(x) = a\cdot0 + b\cdot0 = 0 \ (\bmod{(x-p)})
    [/tex]
    IOW, the original idea is correct.

    LaTeX is hard...
     
    Last edited: Nov 27, 2003
  6. Nov 27, 2003 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, both arcnets and I know what a "factor of a polynomial" is- but that is not normally called a "factor of an equation". Also, you didn't say that "x-p" is a factor, you said "p is a factor".

    I would have interpreted that to mean that, for example, 3 is a factor of 3x2- 6x+ 3= 0 and also a factor of 3x2+ 3x- 9= 0. However, 3 is NOT root of the equation
    3x2- 6x+ 3+ 3x2+ 3x- 9= 0

    What you meant to say was that p was a root of both f(x)= 0 and g(x)= 0. Then, of course, af(p)+ bg(p)= a(0)+ b(0)= 0. (And, in fact, it not required that either p or x-p be factors of f and g- in fact, it is not required that f and g be polynomials.)
     
  7. Nov 29, 2003 #6

    AD

    User Avatar

    Yes, both arcnets and I know what a "factor of a polynomial" is- but that is not normally called a "factor of an equation". Also, you didn't say that "x-p" is a factor, you said "p is a factor".

    Well, I pointed out what the original poster was driving at, I didn't write the question.

    You are correct with what you say, but Denian appears to be from Malaysia, maybe he's not quite au fait with English mathematical terminology.
     
  8. Nov 29, 2003 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    And how will he learn the English terminology if nobody ever tells him what he meant to say?
     
  9. Nov 29, 2003 #8
    well.. actually i know what's the difference btw the factor of the equation and the root of the equation.

    i didnt read the question in the book correctly, and i just type everything from the book.
    but, once i read the question, i quickly make an assumption that
    (x-p) is a factor of the function f and g, and f(p)= g(p) = 0 without thinking any further.

    i dont have much problem with the Eng terminology in math/physics/chem but a lot in Biology. this is the first year that the student of higher school are required to learn Math and Science subjects in English. so, there might be a bit problem, since we used to learn Math & Sc in Malay (for about 10 years)

    anyway, thanks for the explaination
     
  10. Nov 29, 2003 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The crucial point is that saying x= p is a root of f(x)= 0 does NOT mean that (x-p) is a factor of f(x). That is only true if f(x) is a polynomial.
     
  11. Nov 30, 2003 #10
    ?
    an example of what you mean HallsofIvy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Polynomial factorization
  1. Factoring polynomials (Replies: 2)

  2. Taylor Polynomial (Replies: 2)

  3. Polynomial division (Replies: 2)

Loading...