proving a statement about Horner's method

Question

I was reading about Horner's method and then I saw the following statement:

If $f(x)$ is a polynomial such that: $p(x)=\sum_{i=0}^{n}a_{i}x^{i}$,and if we want to >compute the value of this polynomial for $x=x_0$ then defining a sequence of constants : $$b_n=a_n$$ $$b_{n-1}=a_{n-1}+b_{n}x_0$$ implies $$p(x_0)=b_0$$

well I understand this but the problem is here:

Also it can be proven that; $$p\left(x\right)=\left(b_{1}+b_{2}x+...+b_{n}x^{n- 1}\right)\left(x-x_{0}\right)+b_{0}$$ I tried to figure it out, but still I don't know how to prove that.

@hardmath,the source is https://en.m.wikipedia.org/wiki/Horner%27s_method,also it would be really appreciated if you help me with this problem :) — Absurd, Jan 04 '20 at 16:06
@hardmath,yes actually,and using different $x_0$'s generates different $b_i$'s, also if for $x_0$ ,$b_0=0$, implies this $x_0$ is indeed a root of the given polynomial,since $b_0=p(x)$. — Absurd, Jan 04 '20 at 16:21
Okay, one way to see this is by comparing the results of synthetic division, which produces exactly the division of $p(x)$ by $x-x_0$ with remainder $b_0= p(x_0)$ (the evaluation of the polynomial at $x_0$). It will take me a bit to properly craft the $\LaTeX$. — hardmath, Jan 04 '20 at 19:41
Please take a look at this earlier Question, Why does dividing a polynomial by x−a give the same quotient as evaluating it at x=a using synthetic division? If the Answer accepted there is not enough, I will add a new one. . — hardmath, Jan 04 '20 at 19:59
@hardmath,I looked at the page,but can you please add a new one? — Absurd, Jan 06 '20 at 05:58

grand_chat · Answer 1 · 2020-08-19T16:34:15.097

This is just algebra. Expand the RHS and collect the like powers of $x$: $$ \begin{align} &b_0 + (b_1+b_2x+\cdots+b_nx^{n-1})(x-x_0)\\ &=b_0 + b_1x + b_2x^2 +\cdots+b_nx^n-(b_1x_0+b_2x_0x+\cdots+b_nx_0x^{n-1})\\ &=\underbrace{(b_0-b_1x_0)}_{a_0}+ \underbrace{(b_1-b_2x_0)}_{a_1}x+ \underbrace{(b_2-b_3x_0)}_{a_2}x^2+\cdots+ \underbrace{(b_{n-1}-b_nx_0)}_{a_{n-1}}x^{n-1}+\underbrace{b_n}_{a_n}x^n\\ \end{align} $$ In the final step we simply plug in the recursive definitions for $b_0,b_1,\ldots,b_n$, obtaining $p(x)$.

proving a statement about Horner's method

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