# Polynomial factorization

1. Nov 27, 2003

### denian

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. Nov 27, 2003

### arcnets

What is a 'factor of an equation'? I don't understand...

3. Nov 27, 2003

Surely you must know what factorisation of polynomials is, Arcnets.

4. Nov 27, 2003

### arcnets

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.

LaTeX is hard...

Last edited: Nov 27, 2003
5. Nov 27, 2003

### HallsofIvy

Staff Emeritus
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.)

6. Nov 29, 2003

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.

7. Nov 29, 2003

### Hurkyl

Staff Emeritus
And how will he learn the English terminology if nobody ever tells him what he meant to say?

8. Nov 29, 2003

### denian

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

9. Nov 29, 2003

### HallsofIvy

Staff Emeritus
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.

10. Nov 30, 2003

### denian

?
an example of what you mean HallsofIvy.