# Prove that if f(a) = 0, then f(x) = (x-a)g(x)

1. Sep 26, 2009

### nietzsche

1. The problem statement, all variables and given/known data

Prove that if f(a) = 0, then f(x) = (x-a)g(x) where f and g are polynomial functions.

2. Relevant equations

$$x^n-a^n = (x-a)h_n(x)$$
where hn(x) is a polynomial function.

3. The attempt at a solution
\begin{align*} f(x) &= f(x) - f(a)\\ f(x) &= [\lambda_n x^n + \lambda_{n-1} x^{n-1} + ... + \lambda_1 x + \lambda_0] - [\lambda_n a^n + \lambda_{n-1} a^{n-1} + ... + \lambda_1 a + \lambda_0]\\ f(x) &= \lambda_n (x^n-a^n) + \lambda_{n-1} (x^{n-1} - a^{n-1}) +...+ \lambda_1(x-a)+(\lambda_0 - \lambda_0)\\ f(x) &= \lambda_n (x-a)h_n(x) + \lambda_{n-1} (x - a)h_{n-1}(x) +...+ \lambda_1(x-a)\\ f(x) &= (x-a)g(x) \end{align*} where  g(x) = \lambda_nh_n(x) + \lambda_{n-1}h_{n-1}(x) +...+ \lambda_1

Now, I think what I've done here is valid. But I assumed that the relevant equation that I posted is true, which (I think) it is. Can someone tell me if this is an acceptable proof?

Thanks.

Last edited: Sep 26, 2009
2. Sep 26, 2009

### lurflurf

Is f a polynomial?
That is not true in general.
or do you mean in some limit?
consider
f=exp(x)-1
for a polynomial first prove
f(x)=f(a)+(x-a)g(x)

3. Sep 26, 2009

### nietzsche

right, sorry, f is a polynomial. it was the second part of a question, so they had already mentioned that f is a polynomial in the previous part. i'll fix it.

4. Sep 26, 2009

### aPhilosopher

This is correct assuming that you are allowed to use your assumed relevant equation. Did you get it from your book or from somewhere else? If you got it somewhere else, you might want to prove it just to be on the safe side. It's not too difficult.