***Introduction:***
Given two univariate and coprime integer polynomials $f(x), g(x)$, we can always write 
\begin{equation} 
u(x)f(x)+v(x)g(x)=c 
\end{equation}
for a unique pair of integer polynomials $u(x), v(x)$, and a unique positive integer $c$, such that $\deg u<\deg g$ and $\deg v<\deg f$. (This is a consequence of cancellation in the denominators in Bezout's identity, applied to $f$ and $g$ as elements of $\mathbb{Q}[x]$ and $c$ is the $lcm$ of the denominators of the unique pair of rational polynomials determined by Bezout's identity). 


However, the positive integer $c$ need not be the minimal integer satisfying such a relation—there may be integer polynomials $p(x)$ and $ q(x)$ of arbitrary degrees such that 
$$
p(x)f(x)+q(x)g(x)=d
$$
with $d<c$. 


**As an example**, consider $f(x)=2x+1$, $g(x)=2x+17$, for which Bezout's identity gives $c=16$:
$$
\frac{1}{16}(2x+17)-\frac{1}{16}(2x+1)=1 \Leftrightarrow (2x+17)-(2x+1)=16;
$$
while for $p(x)=-(x^4 + 8 x^3 - 4x^2 + 2 x - 1)$, $q(x)=x^4$, we obtain $d=1$:
$$
 (2 x + 17) (x^4)-(2 x + 1) (x^4 + 8 x^3 - 4x^2 + 2 x - 1)=1.
$$
(This example is cited in *[On resultants](http://www.ams.org/journals/proc/1983-089-03/S0002-9939-1983-0715856-2/),* Proc Amer Math Soc 89 (1983) 419-420, G. Myerson.) 


The minimal positive integer $c$ satisfying such a relation $p(x)f(x)+q(x)g(x)=c$, for $p(x), q(x)$ integer polynomials, is by definition called the **reduced resultant** of $f,g$ (see [The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$](https://mathoverflow.net/questions/17501/the-resultant-and-the-ideal-generated-by-two-polynomials-in-mathbbzx/17506#17506)). As I understand it, if $f$ and $g$ are both monic (see [Reduced resultant of monic polynomials](https://mathoverflow.net/questions/227227/reduced-resultant-of-monic-polynomials?noredirect=1&lq=1)), then this minimal positive integer (that is, the reduced resultant), can be found by applying the extended Euclidean algorithm, obtaining the Bezout's identity, and then clearing denominators. 


***Now the question:*** For any pair of coprime, integer polynomials $f$ and $g$, how can we find polynomials $p(x)$ and $q(x)$—of suitable degrees—together with the minimal positive integer $c$, such that
$$
p(x)f(x)+q(x)g(x)=c. 
$$ 


***P.S.:*** Given two coprime integer polynomials $f,g$, by Bezout's identity in $\mathbb{Q}[x]$, we know that all other pairs of rational polynomials, satisfying the relation $A(x)f(x)+B(x)g(x)=1$ are of the form 
$$
A(x)=\frac{1}{c}u(x)+g(x)w(x), \ \ \ B(x)=\frac{1}{c}v(x)-f(x)w(x)
$$
for arbitrary $w(x)\in\mathbb{Q}[x]$. Notice also, that in the preceding example, by long-division we get:
$$
B(x)=q(x)=x^4 + 8 x^3 - 4 x^2 + 2 x - 1=\left(-\frac{1}{16}+\frac{x}{8}-\frac{x^2}{4}+\frac{x^3}{2}\right) (2 x+17) +\frac{1}{16}
$$
and:
$$
A(x)=p(x)=x^4=\left(-\frac{1}{16}+\frac{x}{8}-\frac{x^2}{4}+\frac{x^3}{2}\right) (2 x+1)+\frac{1}{16}
$$
for: $w(x)=\left(-\frac{1}{16}+\frac{x}{8}-\frac{x^2}{4}+\frac{x^3}{2}\right)$. 


Thus, the problem amounts to deciding, whether we can determine a suitable rational polynomial $w(x)$, such that $A(x)$, $B(x)$ have $lcm$ of their denominators less than $c$. 



Reduced resultants and Bezout's identity
First off, it can be seen that the reduced resultant must divide $c$. Indeed, if $u\_1(x)f(x)+v\_1(x)g(x)=c$ and $u\_2(x)f(x)+v\_2(x)g(x)=d$, then from the Bezout identity for integers $c,d$, we can construct the identity $u(x)f(x)+v(x)g(x)=\gcd(c,d)$. Hence, if $d$ is the reduced resultant, then $d=\gcd(c,d)$, i.e. $d\mid c$.


Now, let's see if a prime factor $p|c$ can be cancelled from the r.h.s. of the identity $u(x)f(x)+v(x)g(x)=c$. From polynomials $u(x)$ and $v(x)$, we need to obtain polynomials $cA(x)$ and $cB(x)$ with integer coefficients such that $cA(x)\equiv cB(x)\equiv 0\pmod{p}$. (I follow the notation of OP.)


I'll consider the case $\mathrm{cont}(f)=\mathrm{cont}(g)=1$. 
Then [Gauss' lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)) implies that $cw(x)$ is a polynomial with integer coefficients. Thus, the condition $v(x)-f(x)(cw(x))=cB(x)\equiv 0\pmod{p}$ implies that $f(x)$ divides $v(x)$ modulo $p$. The last condition is necessary and sufficient for cancellation of $p$.
When the divisibility modulo $p$ holds, setting $w(x) = \frac{1}{c}\left(\frac{v(x)}{f(x)}\bmod p\right)$ delivers us the required polynomials $u'(x): = \frac{cA(x)}{p}=\frac{u(x)+g(x)(\frac{v(x)}{f(x)}\bmod p)}p$ and $v'(x): = \frac{cB(x)}{p}=\frac{v(x)-f(x)(\frac{v(x)}{f(x)}\bmod p)}p$, for which
$$u'(x)f(x)+v'(x)g(x) = \frac{c}{p}.$$


So, the reduced resultant can be computed by first computing some $c=u(x)f(x)+v(x)g(x)$ (e.g., with the extended Euclidean algorithm), and then by eliminating prime factors from $c$ (as described above) one by one while this is possible.




---


**Example.** For $f(x)=2x+1$ and $g(x)=2x+17$, we have $c=16$ with $u(x)=-1$ and $v(x)=1$. To check if we can cancel factor $p=2$, we check if $g(x)$ divides $u(x)$ modulo $2$. It does, and we set $c w(x)$ to the quotient, i.e. $cw(x)=1$, to get new polynomials $u\_2(x):=\frac{u(x) + g(x)}2 = x+8$ and $v\_2(x):=\frac{v(x)-f(x)}2=-x$. It is easy to check that
$$u\_2(x)f(x) + v\_2(x)g(x) = 8,$$
i.e. the factor $2$ is cancelled in the r.h.s.


Continuing this way, we can cancel another factor of $2$ by setting $cw(x)=x$ to get $u\_3(x):=\frac{u\_2(x)+g(x)x}2 = x^2 + 9x + 4$ and $v\_3(x):=\frac{v\_2(x)-f(x)x}2 = -x^2 - x$ with
$$u\_3(x)f(x) + v\_3(x)g(x) = 4,$$
and so on.


