# Proof of Remainder Theorem for polynomials

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

Proof: (Prove by strong induction)

Base case $n = 1: f(x) = cx + d$, then let $g(x) = c, b = d + ac$. That is: $f(x) = cx + d = c(x-a) + (d+ac)$

Assume for polynomials with degree $\le k$

To show: $k + 1$ that $f(x) = a_{k+1}x^{k+1} + \ldots + a_1x + a_0$.

Consider $h(x) = f(x) - a_{k+1}x^{k}(x-a)$ has a degree $\le k$

Then: $f(x) - a_{k+1}x^{k}(x-a) = (x-a)g(x) + b$

Finally: $f(x) = (x-a)(g(x) + x^{k}a_{k+1}) + b$

Thanks.

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I've edited for formatting, but I don't know what you really meant when you wrote $h(x)=f(x)-ak+1(x-a)$ so I left that alone. I'm pretty sure, though, that it's not what you meant to write. – Gerry Myerson Sep 19 '12 at 13:29
That looks better, but are you sure it isn't supposed to be $f(x)-a_{k+1}x^k(x-a)$? Now, that would have degree at most $k$. – Gerry Myerson Sep 19 '12 at 13:33
Maybe that's right, then its typo in the book. I will complete the proof above and please check if it is right. – mathnoob Sep 19 '12 at 13:40
Looks good to me. – Gerry Myerson Sep 19 '12 at 13:47
I came across this too, the proof seems to be given incorrectly in the solutions manual for Spivak's Calculus (Chapter 3, Question 7). – Speedy Nov 25 '13 at 15:23

Your proof is correct (but the base case should be $\rm\:n = 0).\:$ Here's a simpler proof: by shifting $\rm\:x\to x+a,\:$ we reduce to the case $\rm\:a=0,\:$ then $\rm\:f(x)\, =\, f_ 0 + x\,(f_1 + f_2 x + \cdots + f_n x^{n-1})\,\$ QED