Can $\frac{P(x)s(x)+Q(x)}{R(x)}$ be a polynomial for some polynomial $s(x)$? Given some polynomials $P(x), Q(x), R(x), \operatorname{gcd}(P(x), R(x)) = 1$ does there exist some polynomial $s(x)$ so
$$
\frac{P(x)s(x) + Q(x)}{R(x)}
$$
is also a polynomial? If so how do I construct one?
A concrete example I'm working with is
$$
R(x) = 480 (12 x^2-1)\\
Q(x) = -48 x^5+280 x^3-259 x\\
P(x) = 1920 x^3+3360 x
$$
 A: As was noted in the comments: requiring that $\dfrac{P(x)s(x)+Q(x)}{R(x)}$ be equal to a polynomial (which we'll call $t(x)$ here) is equivalent to requiring that $P(x)s(x)+Q(x)=R(x)t(x)$, or that $P(x)s(x)-R(x)t(x)=Q(x)$.  Now the aforementioned Bezout's identity kicks in; since $\gcd(P(x),R(x))=1$ then there are even polynomials $u$ and $v$ with $P(x)u(x)-R(x)v(x)=1$; you can then take $s(x)=u(x)Q(x)$, $t(x)=v(x)Q(x)$.  Note that the maximum degree here will be $\deg(Q)+\max(\deg(P),\deg(R))-1$; if the degree of $Q$ is less than the degree of the other two polynomials and you're looking for a bound along the lines of $\max(\deg(P),\deg(R))-1$ or if you want $\gcd(s(x),Q(x))=1$ as well, then those are substantially harder and I'm not entirely certain a solution exists.
A: Here's one possible solution: to make
$$
\frac{P(x)s(x) + Q(x)}{R(x)}
$$
a polynomial we need to ensure that every root $x_i$ of the denominator $R(x_i) = 0$ is also a root of the numerator, so
$$
P(x_i) s(x_i) + Q(x_i) = 0\\
s(x_i) = -\frac{Q(x_i)}{P(x_i)}
$$
and since $\operatorname{gcd}(P, R) = 1$, $P(x_i)$ cannot be zero. This is a classical interpolation problem and can be easily solved.
