Proving Polynomial Proofs: Using the Expansion Method

[courtrigrad]

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?

 

[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)?

 

[courtrigrad]

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

 

[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)?

 

[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.