Basic Mathematics. Problem with polynomial function proof

Question

Let $f$ be a polynomial of degree $\leqslant n$ and let $c$ be a root. Then there exists a polynomial $g$ of degree $\leqslant n-1$ such that for numbers $x$ we have $$ f(x)=(x-c)g(x)$$

Proof: Write

$$ f(x)=a_{0}+a_{1}x+a_{2}x^{2}+ ... + a_{n}x^{n} $$

Then substitute the value

$$ x = (x-c) + c$$

for each x.

Here is the point where I get stuck, as after substituting $x$, the author derives

$$ f(x)=b_{0}+b_{1}(x-c)+b_{2}(x-c)^{2}+ ... + b_{n}(x-c)^{n} $$

I cannot understand how he gets the above function as it looks like that he keeps only the term $(x-c)^n$ of every expansion $ ((x-c) + c)^n$.

Any hints?

Answer 1

If you divide $f$ with $x-c$ then you get some $g$ of degree $n-1$ and residue $r$ (constant polynomial) such that $$f(x) = g(x)(x-c)+r$$

Now since $c$ is a root of $f$ we have $$0=f(c) = g(c)(c-c) +r\implies r=0$$

so we have $$ f(x)= (x-c)g(x)$$

Answer 2

By the Binomial theorem,

$$((x-c)+c)^m=\sum_{k=0}^m\binom mkc^k(x-c)^{m-k}=c^{m}+(x-c)P(x-c)$$ where $P$ is some polynomial of degree $m-1$.

Then

$$F(x)=\sum_{j=0}^nf_jx^j=\sum_{j=0}^nf_j((x-c)+c)^j=\sum_{j=0}^nf_jc^j+(x-c)Q(x-c)$$ where $Q$ is another polynomial, of degree $n-1$.

As $c$ is a root of $f$, the first term vanishes.

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E.g.,

$$\begin{align}F(x)&=px^2+qx+r \\&=p((x-c)+c)^2+q((x-c)+c)+r \\&=p(x-c)^2+2pc(x-c)+pc^2+q(x-c)+qc+r \\&=(x-c)(p(x-c)+2pc+q)+pc^2+qc+r \\&=(x-c)(p(x-c)+2pc+q).\end{align}$$