Using Horner's Method I'm trying to evaluate a polynomial recursively using Horner's method.
It's rather simple when I have every value of $x$ (like: $x+x^2+x^3...$), but what if I'm missing some of those? Example: $-6+20x-10x^2+2x^4-7x^5+6x^7$.
I would also appreciate it if someone could explain the method in more detail, I've used the description listed here but would like some more explanation.
 A: Horner's method or form is also sometimes called nested form.  You can think of it as starting with the whole polynomial $$6x^7-7x^5+2x^4-10x^2+20x-6,$$ setting aside the constant term (if it's zero, you can set aside a zero here) and factoring out an $x$ from the remaining terms $$(6x^6-7x^4+2x^3-10x+20)x-6,$$ and then repeating the process for the innermost polynomial (the part that isn't yet fully in nested form:
$$((6x^5-7x^3+2x^2-10)x+20)x-6$$
$$(((6x^4-7x^2+2x)x-10)x+20)x-6$$
$$((((6x^3-7x+2)x+0)x-10)x+20)x-6$$
$$(((((6x^2-7)x+2)x+0)x-10)x+20)x-6$$
$$((((((6x)x-7)x+2)x+0)x-10)x+20)x-6$$
$$(((((((6)x+0)x-7)x+2)x+0)x-10)x+20)x-6$$
Alternately, if your polynomial is $$p(x)=\sum_{k=0}^{n}a_kx^k=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$ you can think of $p(x)$ in nested form as $p_0$ where $p_n=a_n$ and $p_{k-1}=p_kx+a_{k-1}$.  That is, start with the leading coefficient ($a_n$).  Multiply by $x$ and add the next coefficient (adding zero to fill in for "missing" powers of $x$), repeating until you've added the constant term ($a_0$). Back to the same example:
$$\begin{align}
p_7&=6\\
p_6&=(6)x+0\\
p_5&=((6)x+0)x-7\\
p_4&=(((6)x+0)x-7)+2\\
p_3&=((((6)x+0)x-7)+2)x+0\\
p_2&=(((((6)x+0)x-7)+2)x+0)-10\\
p_1&=((((((6)x+0)x-7)+2)x+0)-10)+20\\
p_0&=(((((((6)x+0)x-7)+2)x+0)-10)+20)-6
\end{align}$$
A: You can also carry it out in a synthetic division table.  Suppose you want to evaluate $f(x) = x^4 - 3x^2 + x - 5$ for $x = 3$. Set up a table like this


              1    0     -3    1     5
          3
             -------------------------
              1

Now multiply the number on the bottom and total as follows.


              1    0     -3    1     5
          3        3
             -------------------------
              1    3

Work your way across in this manner.


              1    0     -3    1     -5
          3        3      9   18    57
             -------------------------
              1    3      6   19    52

We have $f(3) = 52$.  Let's run a check
$$ f(3) = 81 -3*9 + 3 - 5 = 54 - 2 = 52.$$
$$ f(3) = 54 - 2 = 52.$$
This is a clean, tabular way to see Horner's method work.  
A: $6x^7-7x^5+2x^4-10x^2+20x-6$ =
$6x^7+0x^6-7x^5+2x^4+0x^3-10x^2+20x-6$
The method is essentially start with the coefficient of the highest power, multiply by $x$ and add the next coefficient. Stop when you add the constant coefficient.
So steps in the iteration go:
$6$
$6x[+0]$
$6x^2-7$
$6x^3-7x+2$
$6x^4-7x^2+2x[+0]$
$6x^5-7x^3+2x^2-10$
$6x^6-7x^4+2x^3-10x+20$
$6x^7-7x^5+2x^4-10x^2+20x-6$
Trust this helps
