Remainder of division of $r(x)h(x)$ by $g(x)h(x)$ (that is known to have degree less than degree of $g(x)$) is equal to that of $r(x)$ by $g(x)$ Is the conclusion below correct?


  Remainder of division of $r(x)h(x)$ by $g(x)h(x)$ (that is known to have degree less than degree of $g(x)$) is equal to remainder of division of $r(x)$ by $g(x)$ ?
No.  Division of a polynomial $P(x)$ by $D(x)$ takes the form $P(x)=Q(x)D(x)+R(x)$, where $R$ is the remainder and satisfies $\deg R&lt;\deg Q$.  Let $r(x)=g(x)Q(x)+R(x)$, where $\deg R&lt;\deg g$.  Then $r(x)h(x)=g(x)h(x)Q(x)+R(x)h(x)$, where the remainder satisfies $$\deg (R\cdot h)=\deg R+\deg h&lt;\deg g+\deg h=\deg(g\cdot h)$$  In conclusion, the remainder is multiplied by $h(x)$, but is not the same.
While to show that something is true it is necessary a full demonstration, to verify that it is wrong one example is enough, so let's get $$r=3, h=5, g=2$$. $$r\%g=3\%2=1$$ $$(rh)\%(gh)=15\%10=5$$. This shows that the assumption is false, infact 1 is different from 5!

Int the previous

This is the good and the bad about math... a lot of work to demonstrate something and a simple example to destroy the theory ;)