prove that ,for any polynomial function $f$ ,f can be written as $f(x)=(x-a)g(x)+b$ 
prove that for any polynomial function $f$ ,and any number $a $,there is a polynomial function $g$ ,and a number $b$,such that $f(x)=(x-a)g(x)+b$

this question from spivak calculus book.
and this is the answer:
if the degree of $f$ is $1$ then f is on the form
$f(x)=cx+d=c(x-a)+(d+ax)$
suppose the result is true for  the polynomials of degree $<k$.if $f$ has degree $k+1$,then $f$ has the form
$f(x)=a_{k+1}x^{k+1}+...+a_1x +a_0$
now the polynomial function $h(x)=f(x)-a_{k+1}(x-a) $ has degree $\leq k$ (*),so we can write
$f(x)-a_{k+1}(x-1)=(x-a)g(x)+b$ (**)
or $f(x)=(x-a)(g(x)+a_{k+1})+b$.
i can't understand the step (*),i think $h(x)$ has degree $\leq k$ just if $h(x)=f(x)-a_{k+1}(x^{k+1}-a) $, not when $h(x)=f(x)-a_{k+1}(x-a) $
and also in the step (**) i think it must be $f(x)-a_{k+1}(x^{k+1}-a)=(x-a)g(x)+b$ .
so i did not  understand this proof ,so can you explain it more?
 A: Every monomial $x^n$ can be expressed as a polynomial of $X=x-a$ of same degree by the binomial formula.
$x^n=(X+a)^n=\sum\limits_{i=0}^n\binom{n}{i}X^ia^{n-i}$ and by linearity any polynomial in $x$ can be rewritten as a poynomial in $X$ of the same degree.
And reciprocally for $X^n=(x-a)^n$, polynomials in $X$ can be rewritten as polynomials in $x$ of the same degree.
Expressed differently this means that $\Big(1,x,x^2,x^3,\cdots\Big)$ and $\Big(1,(x-a),(x-a)^2,(x-a)^3,\cdots\Big)$ are both basis for polynomial vector space.
The results follows immediately since:
$$\begin{align}p(x)&=\underbrace{u_0}_b+u_1(x-a)+u_2(x-a)^2+\cdots\\
&=(x-a)\underbrace{\Big(u_1+u_2(x-a)+\cdots\Big)}_{g(x)}+b\end{align}$$
A: If one insists on proving the statement by induction, the induction step could be performed as follows. For $\deg f=0$ the conclusion is obvious, with $g=0.$ Assume the conclusion holds for polynomials of degree less than or equal $k.$ Assume $\deg f=k+1.$ Hence
$$f(x)=a_{k+1}x^{k+1}+\ldots +a_1x+a_0$$ Then
$$h(x):=f(x)-a_{k+1}x^k(x-a)$$ satisfies $\deg g\le k.$ By induction hypothesis we have
$$f(x)-a_{k+1}x^k(x-a)=(x-a)g(x)+b$$ for a polynomiall $g(x).$ Thus
$$f(x)=(x-a)[a_{k+1}x^k+g(x)]+b$$
