Given two polynomials, determine two other degrees through polynomial division

Question

I read in a proof, where it says something in this direction

Divide any polynomial $Q$ of degree $2n-2$ by $P$ (of degree $n$) and get an equation $Q=SP+R$ with $S$ and $R$ polynomials of degree at most $n-2$ and $n-1$, respectively.

If two polynomials $P,Q$ are given with $\deg Q=2n-2$ and $\deg P=n$, it is well-known that we can write $Q=SP+R$ for some unique polynomials of $S$ and $R$ with $\deg R<\deg S$. The question is, how do you see that $\deg S\leq n-2$ and $\deg R\leq n-1$, as stated above?

Answer 1

Sum of degrees of $S$ and $P$ must equal to the degree of $Q$ so $$\deg S +n = 2n-2\implies \deg S = n-2$$

Since $\deg R < \deg P=n$ we have $\deg R\leq n-1$.

Answer 2

an example....

$$ \left( x^{8} + 2 x^{7} + 3 x^{6} + 4 x^{5} + 5 x^{4} + 6 x^{3} + 7 x^{2} + 8 x + 9 \right) $$

$$ \left( x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \right) $$

$$ \left( x^{8} + 2 x^{7} + 3 x^{6} + 4 x^{5} + 5 x^{4} + 6 x^{3} + 7 x^{2} + 8 x + 9 \right) = \left( x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( x^{3} + x^{2} + x + 1 \right) } + \left( x^{4} + 2 x^{3} + 4 x^{2} + 6 x + 8 \right) $$