- #1

nietzsche

- 186

- 0

## Homework Statement

Prove that for any polynomial function f, and any number a, there is a polynomial function g, and a number b, such that f(x) = (x-a)g(x) + b for all x.

## Homework Equations

## The Attempt at a Solution

Base Case: Let f(x) be a polynomial of degree n = 1. Then:

[tex]

\begin{align*}

f(x) &= c_1x+c_0\\

f(x) &= c_1x - c_1a + c_1a + c_0\\

f(x) &= (x-a)c_1 + c_1a + c_0\\

f(x) &= (x-a)g(x) + b

\end{align*}

[/tex]

where g(x) = c

_{1}and b = (c

_{1}a + c

_{0}).

Let P(n) be the statement that every polynomial function of degree n can be written as f(x) = (x-a)g(x) + b.

Take P(n+1) and assume P(n) is true:

[tex]

\begin{align*}

f(x) &= c_{n+1}x^{n+1} + c_{n}x^{n}+...+c_1x + c_0\\

f(x) &= c_{n+1}x^{n+1} + (x-a)q(x) + k_0\\

f(x) &= c_{n+1}x^{n+1} - c_{n+1}x^na + c_{n+1}x^na + (x-a)q(x) + k_0\\

f(x) &= (x-a)(c_{n+1}x^n) + c_{n+1}x^na + (x-a)q(x) + k_0\\

f(x) &= (x-a)(c_{n+1}x^n) + (x-a)r(x) + k_1 + (x-a)q(x) + k_0\\

f(x) &= (x-a)(c_{n+1}x^n + r(x) + q(x)) + k_1 + k_0\\

f(x) &= (x-a)g(x) + b

\end{align*}

[/tex]

where g(x) = c

_{n+1}x

^{n}+ r(x) + q(x)

and b = k

_{1}+ k

_{0}.

I think it's right, but I'm always so unsure of myself when it comes to induction.