If you know for a fact that your points lie on a polynomial of degree 5, you can use finite differences. This is especially easy when the $x$-values are in arithmetic progression, as in your case. Write the $y$-values:
175 125 100 125 0 -125 -100 -125 -175
Then write the difference between each number and the next:
-50 -25 25 -125 -125 25 -25 -50
Again:
25 50 -150 0 150 -50 -25
Again:
25 -200 150 150 -200 25
Again:
-225 350 0 -350 225
And once more:
575 -350 -350 575
Now if your points were really from a polynomial of degree 5, that last line would have been constant, but it's not, so they're not. But after all, you said they were estimated points - they still might be close to some polynomial of degree 5. To find the polynomial of degree 5 that comes closest to your points, there is a method called least squares, and there are many expositions of it on the web.
EDIT: Here's a start on least squares:
You want a polynomial $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ such that $p(800)=175$, $p(600)=125$, etc. We already know that's impossible, so we settle for making all the numbers $p(800)-175$, $p(600)-125$, etc., small. In fact, we form the quantity $$(p(800)-175)^2+(p(600)-125)^2+\cdots+(p(-800)-175)^2$$ and we try to minimize it. This quantity is a function of the 6 variables $a,b,c,d,e,f$, and there are standard calculus techniques for minimizing such a function. It gets very messy, but fortunately you don't actually have to do it; someone has done it for you, in the most general case, and found that you can write down a very simple matrix equation which you can solve for the unknowns $a,b,c,d,e,f$. And that's what you'll find if you search for least squares polynomial fit.
You can do this most easily with matrices, and a quick Google search shows that matrix multiplication can be done in Excel, but it can be a bit of a pain, especially with so many points.
Let $A$ be the matrix created from the system of equations that result from plugging each point into $f(x)$ and $y$ be the matrix containing the values of $f(x)$. So given a set of points $\{(x_1, y_1),(x_2, y_2),...,(x_n, y_n)\}$, if you want a polynomial fit of degree $N$ (with $N>n$), you'd have the following matrices:
$A = \begin{bmatrix} x_1^{N} & x_1^{N-1} & x_1^{N-2} & ... & 1 \\ x_2^{N} & x_2^{N-1} & x_2^{N-2} & ... & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_n^{N} & x_n^{N-1} & x_n^{N-2} & ... & 1 \end{bmatrix}$
$Y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$
Now take set the matrix $\beta$ as
$$\beta = (A^T A)^{-1} A^T Y$$ whose elements are the coefficients of the $N$th degree polynomial corresponding to the set of points.
Best Answer
The rule being discussed in the video applies when the $x$ values are equally spaced. If you have a linear graph, then equally spaced $x$ values will correspond to $y$ values that are also equally spaced (with different spacing, usually), so the differences in those $y$ values will be constant.
In your problem the $x$ values are not equally spaced. This is the reason why you reach the wrong conclusion when applying the methods of the video.