For the polynomial divison we can see it as;

#(-x^5+7x^3-x) : (x^3-x^2+1) = #

So basically, what we want is to get rid of #(-x^5+7x^3-x)# here with something we can multiply on #(x^3-x^2+1)#.

We can start out with focusing on the first parts of the two, #(-x^5) : (x^3)#. So what do we need to multiply #(x^3)# with here in order to achieve #-x^5#? The answer is #-x^2#, because #x^3*(-x^2)=-x^5#.

So, #-x^2# will be our first part for the polynomial long divison. Now though, we cannot just stop at multiplying #-x^2# with the first part of #(x^3-x^2+1)#. We have to do it for each of the operands.

In that case, our first chosen operand will give us the result of;

#x^3*(-x^2)-x^2*(-x^2)+1*(-x^2)#. Though there is one extra thing, there's always a #-# (minus) operator before the divison. So the notation would actually be something like,

#(-x^5+7x^3-x) : (x^3-x^2+1) = color(red)(-x^2)#

#-(-x^5+x^4-x^2)#

Which will give us,

#(-x^4+7x^3+x^2-x):(x^3-x^2+1)#

A little notice here is that any operand that isn't taken out by the divison is carried on. That is untill we cannot do any divison. Meaning that we cannot find anything to multiply #(x^3-x^2+1)# with in order to take out any elements from the left side.

I will continue with the notation now,

#(-x^4+7x^3+x^2-x):(x^3-x^2+1) = color(red)(-x)#

#-(-x^4+x^3-x)#

#=>(6x^3+x^2) : (x^3-x^2+1)#

#(6x^3+x^2) : (x^3-x^2+1) = color(red)(6)#

#-(6x^3-6x^2+6)#

#=>(7x^2+6) : (x^3-x^2+1)#

It's a stop here. Because #(x^3-x^2+1)# contains a #x^3# and there is nothing on the left side that will need something #x^3#. We will then have our answer as;

#=-x^2-x+6+(7x^2-6)/(x^3-x^2+1)#