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denian
Nov27-03, 08:58 AM
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.

arcnets
Nov27-03, 02:05 PM
What is a 'factor of an equation'? I don't understand...

AD
Nov27-03, 03:54 PM
Surely you must know what factorisation (http://mathworld.wolfram.com/PolynomialFactorization.html) of polynomials is, Arcnets.

arcnets
Nov27-03, 04:07 PM
Yeah, OK. So f(x) and g(x) share the factor (x-p), right?

f(x) = 0 \ (\bmod{(x-p)}) \ \wedge \ g(x) = 0 \ (\bmod{(x-p)})


\Rightarrow


af(x) + bg(x) = a\cdot0 + b\cdot0 = 0 \ (\bmod{(x-p)})

IOW, the original idea is correct. [:D]

LaTeX is hard...

HallsofIvy
Nov27-03, 06:45 PM
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.)

AD
Nov29-03, 07:49 PM
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.

Hurkyl
Nov29-03, 08:02 PM
And how will he learn the English terminology if nobody ever tells him what he meant to say?

denian
Nov29-03, 08:29 PM
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

HallsofIvy
Nov29-03, 09:41 PM
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.

denian
Nov30-03, 12:36 AM
?
an example of what you mean HallsofIvy.