Descartes and understanding analytical geometry this is my first post and I hope I'm going about this the right way. I am currently trying to understand how Descartes made sense of algebra using geometry. So i found this lovely pdf: https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/1996/0025570x.di021189.02p0130a.pdf
On page 6 it describes how he geometrically would show his algebra to give it substance. Moving away from the Greek ideas and demonstrating that two lengths could be multiplied to equal another length, instead of an area, x^2=a. This explanation I'm content with. However I do not grasp how Descartes could show this principle to be true for x^5, x^10 or any arbitrary order. Can someone help me or give me links to other information?
Thanks for all considerations 
(My math level is second year of engineering if that is relevant)
 A: You acknowledge in a comment that you just want to construct $x^n$. Well, given a segment $\overline{OP_0}$ of length $1$, and a segment $\overline{OP_1}$ of length $p$, then one can construct segments $\overline{OP_n}$ of length $p^n$ with a simple spiral of perpendicular lines:

The construction leverages the proportions of the similar right triangles $\triangle OP_nP_{n-1}$ and $\triangle OP_{n-1}P_{n-2}$ for which we have
$$\frac{|\overline{OP_n}|}{|\overline{OP_{n-1}|}} = \frac{|\overline{OP_{n-1}}|}{|OP_{n-2}|} \qquad\to\qquad |\overline{OP_n}| = \frac{\;|\overline{OP_{n-1}}|^2}{|OP_{n-2}|}$$
If the "earlier" segments obey the relation $|\overline{OP_k}| = p^k$ (which they certainly do for $k = 0$ and $k=1$), then we have
$$|\overline{OP_n}| = \frac{\left(p^{n-1}\right)^2}{p^{n-2}} = \frac{p^{2n-2}}{p^{n-2}} = p^n$$
so that segment $\overline{OP_n}$ satisfies the relation, too. By induction, we get any (non-negative integer) power of $p$ we like. (Note that we can continue the spiral "backwards" from $P_0$ to get the negative-integer powers $p^{-1}$, $p^{-2}$, $\dots$ .)
Observe that having an auxiliary segment of length $1$, against which to compare the segment of length $p$, is crucial to this construction ... and to any construction, really. If we have a segment of length $p$, and nothing else, then there's no telling what a segment of length $p^2$ would look like: If $p$, compared to some "hidden" unit, turns out to be, say, $2$, or $1/2$, or $1$, then the segment of length $p^2$ would either be twice as long ($p^2 = 4 = 2p$), half as long ($p^2 = 1/4 = p/2$), or just as long ($p^2 = 1 = p$) as the original segment. (We don't have this problem when turning a segment of length $p$ into a square of area $p^2$, because any "hidden" unit of length is automagically compatible with the corresponding "hidden" unit of area to make this work.)
I don't know that this helps with understanding the Extended Pappus Four-Line Problem, however.
