Algebra Query

[Dale10101]

Is it true that given a rational polynomial, f(x), that in order for f(x) = 0 to be true for a value x = a, it must then be true that there must exist an identity of f(x) such that (x-a) is extractable as a factor, that is, that f(x) = (x-a)(R) where R is some rational polynomial.

For example:

\[\begin{array}{l}
f(x) = \frac{x}{{x - 3}} - 2\\
x = 6\\
f(6) = \frac{6}{{6 - 3}} - 2 = 0
\end{array}\]

implies a B such that

\[f(6) = (x - 6)(B)\]

I know that it is true in this case and in fact (after a mild algebraic trick that I had not seen in algebra text books):

\[f(x) = \frac{x}{{x - 3}} - 2 = \frac{{(x - 6)}}{{(3 - x)}}\]

but is this case true in general, if not a simple counter example would be appreciated (I am prepared to be slapped by the the obvious, but anyway). If the general case is true for rational polynomials is it true when you include exponential, logarithmic and trigonometric expressions.

I was led into this query while trying to understand how proofs of calculus limits work. Thanks.

 

[daon2]

Yes, if \(\displaystyle f(x)=\dfrac{p(x)}{q(x)}\) then \(\displaystyle f(r)=0\) if and only if \(\displaystyle q(r)\neq 0\) and \(\displaystyle p(r)=0\)

And we know that since \(\displaystyle p(x)\) is a plynomial, \(\displaystyle p(x)=(x-r)g(x)\)

So \(\displaystyle f(x) = (x-r)\cdot\dfrac{g(x)}{q(x)}\).

 

[stapel]

Try taking a look at the Factor Theorem and / or the Remainder Theorem.

 

[Dale10101]

Thanks

The Remainder and Factor theorem seems straight forward enough, even the proofs are short and simple, still ... something seems surprising about the result, still cogitating ...

Much headache later (am I a masochist, ):

I think the fact that one can extend the meaning of the “Division Algorithm” from arithmetic to polynomials that is surprising.

That is, one can easily see division as repeated subtraction with the likelihood of a remainder and thus that a dividend can be re-expressed as the product of divisor and a quotient plus a remainder (zero or otherwise), but it doesen’t seem obvious that a polynomial expression in x can be re-expressed as the product of a linear polynomial and some other polynomial plus, possibly, a remainder … FOR ALL VALUES OF X within the polynomials domain… that last seems the surprising thing, must look into the proof of the division algorithm.

Well, no comment necessary, thanks for your input.

 

[JeffM]

The Remainder and Factor theorem seems straight forward enough, even the proofs are short and simple, still ... something seems surprising about the result, still cogitating ...

Much headache later (am I a masochist, ):

I think the fact that one can extend the meaning of the “Division Algorithm” from arithmetic to polynomials that is surprising.

That is, one can easily see division as repeated subtraction with the likelihood of a remainder and thus that a dividend can be re-expressed as the product of divisor and a quotient plus a remainder (zero or otherwise), but it doesen’t seem obvious that a polynomial expression in x can be re-expressed as the product of a linear polynomial and some other polynomial plus, possibly, a remainder … FOR ALL VALUES OF X within the polynomials domain… that last seems the surprising thing, must look into the proof of the division algorithm.

Well, no comment necessary, thanks for your input.

Fundamental theorem of algebra: Any polynomial of degree n with real coefficients can be expressed as a product of a linear term with real coefficients and (n - 1)/2 quadratic terms with real coefficients if n is odd and as a product of (n/2) quadratic terms with real coefficients if n is even. The proof is not easy.

I suspect that you are asking several questions at once.

 

[Dale10101]

Right

You are probably right JeffM, I am reading that Gauss proved the "Fundamental Theorem of Algebra" in 1799 and that it was not a forgone conclusion.

I doubt I am equipped to pursue an understanding of that proof at this point, no doubt I must focus on simpler conceptual problems as well as continue to work on the standard text book problems while just peeking through the knotholes in the great of fence of ignorance :sad: (Oh brother!, am laughing.) Good day.:grin: