A polynomial of degree (or order) $n$ is given by
$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0.$$The numbers $\{a_i\}_{i=0}^n$ are called the coefficients.
A polynomial $P(x)$ with real or complex coefficients has $n$ (possibly repeated and possibly complex) roots. Furthermore, for some (possibly complex) values $x_1,x_2,\ldots,x_k$, and integers $m_1,m_2,\ldots,m_k$
$$P(x) = a_n (x-x_1)^{m_1} (x-x_2)^{m_2} \cdots (x-x_k)^{m_k}, \quad \sum_{j=1}^k m_k = n.$$The roots of a polynomial equation, $P(x) = 0$, could be found by Bisection or Fixed-point iteration. Both require computing $P(x)$ at various value of $x$, which can be computationally intensive because there are so many multiplications and additions in a (high-degree) polynomial.
Better convergence is achieved by using the Newton method, but the trade-off is that there are even more multiplications and divisions because not only $P(x)$ but also $P'(x)$ needs to be computed.
In Lec 2, we realized that nesting a polynomial reduced the number of operations needed to compute it. Now we introduce a systematic method, Horner's method, that effectively does computes a nested representation of a polynomial.
Let
$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0.$$Define $b_n = a_n$ and
$$ b_k = a_k + b_{k+1}x_0, \quad \text{for } k = n-1, n-2,\ldots,1,0.$$Then $b_0 = P(x_0)$. Moreover, if
$$ Q(x) = b_n x^{n-1} + b_{n-1} x^{n-2} + \cdots + b_2 x + b_1,$$then
$$ P(x) = (x-x_0) Q(x) + b_0.$$If we can prove the last claim then $P(x_0) = b_0$. So, consider
$$ (x-x_0) Q(x) + b_0\\ = x Q(x) - x_0 Q(x) + b_0\\ = b_n x^{n} + b_{n-1} x^{n-1} + \cdots + b_2 x^2 + b_1 x\\ - b_n x_0 x^{n-1} - b_{n-1} x_0 x^{n-2} - \cdots - b_2 x_0 x - b_1 x_0 + b_0$$Collecting the powers, we find
This we have an iterative and computationally efficient method for computing $P(x_0)$.
Now, recall that we are interested in speeding up Newtons method, which calls for not only $P(x_0)$, but also $P'(x_0)$.
But here's the clever part:
$$ P'(x) = Q(x) + (x-x_0)Q'(x) \quad\Rightarrow\quad P'(x_0) = Q(x_0).$$