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

[nietzsche]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

$$\begin{align*}<br /> f(x) &= f(x) - f(a)\\<br /> f(x) &= [\lambda_n x^n + \lambda_{n-1} x^{n-1} + ... + \lambda_1 x + \lambda_0] - [\lambda_n a^n + \lambda_{n-1} a^{n-1} + ... + \lambda_1 a + \lambda_0]\\<br /> f(x) &= \lambda_n (x^n-a^n) + \lambda_{n-1} (x^{n-1} - a^{n-1}) +...+ \lambda_1(x-a)+(\lambda_0 - \lambda_0)\\<br /> f(x) &= \lambda_n (x-a)h_n(x) + \lambda_{n-1} (x - a)h_{n-1}(x) +...+ \lambda_1(x-a)\\<br /> f(x) &= (x-a)g(x)<br /> \end{align*}<br /> where g(x) = \lambda_nh_n(x) + \lambda_{n-1}h_{n-1}(x) +...+ \lambda_1$$

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.

 

[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)

 

[nietzsche]

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)

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.

 

[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.