# Homework Help: Polynomial Proofs

1. May 29, 2006

If $$n \geq 1$$ and $$f(a) = 0$$ for some real a , then $$f(x) = (x-a)h(x)$$, where h is a polynomial of degree $$n-1$$. So:

$$f(a) = \sum_{k=0}^{n} c_{k}a^{k} = c_{0} + c_{1}a + c_{2}a^{2} + ... + c_{n}a^{n} = 0$$. In a hint it says to consider $$p(x) = f(x+a)$$. So I expanded that and got: $$c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n}$$. So how do I use this to prove the above statement?

2. May 29, 2006

### AKG

If there exists a polynomial h(x) such that f(x) = (x-a)h(x), we say that (x-a) is a factor of f(x). In general, if f(x) and g(x) are any two polynomials, then if there exists h(x) such that f(x) = g(x)h(x), we say that g(x) is a factor of f(x). So the problem asks you to prove that if f(a) = 0, then (x-a) is a factor of f(x). Do you see that it suffices to prove that x is a factor of p(x), given that p(x) is defined as f(x+a)?

Well you know that f(a) = 0, which gives:

$$c_{0} + c_{1}a + c_{2}a^{2} + ... + c_{n}a^{n} = 0$$

And you've written out p(x) as:

$$c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n}$$

Can you put these two facts together to show that x divides p(x), i.e. that x is a factor of p(x)?

3. May 29, 2006

Well $$c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n} = c_{0} + c_{1}x+c_{1}a + c_{2}x^{2} + 2c_{2}xa + c_{2}a^{2} + ... + c_{n}(x+a)^{n}$$. I know that we can factor out the $$f(a)$$ and set it equal to 0. But then what is the new expression. Also could you explain why it suffices to prove that x is a factor of $$p(x)$$ if $$p(x) = f(x+a)$$?

Thanks

4. May 29, 2006

### Nimz

You want to show that (x-a) is a factor of f(x). Do you see how this is related to showing that x is a factor of p(x)=f(x+a)? Or perhaps showing that t is a factor of f(t+a)?

5. May 30, 2006

### e(ho0n3

When I first looked at this, I didn't understand it either. But after some deep thinking, it started making sense.

If (x - a) is a factor of f(x), the x is a factor of p(x). Why? Because if we assume that f(x) = (x - a) h(x) for some h(x), then p(x) = f(x + a) = (x + a - a)h(x) = x h(x).

Do you understand now?

I'll give you a hint to prove that x is a factor of p(x): p(x) = f(x + a) = f(x + a) - f(a) since f(a) = 0.