suppose you have two integers n,m with no common factors except 1 and -1, and pose the same question. then you can find two integers a and b such that an+bm = 1. (This is due to Euclid, more than one thousand years before bezout, and his argument works also for relatively prime polynomials. The theorem of bezout relates to two polynomials in two variables and asks how many common points their curves of zeroes can have in the plane.)
anyway all you have to do is divide the samller one by the alreger one, and then repeat this with the remainder, until finally you get down to one.
this is called the euclidean algorithm.
example: take 39 and 28. divide 39 by 28 get 39 = 1(28) +11. then divide 28 by 11, and get 28 = 2(11) + 6. then divide 11 by 6 and get 11 = 1(6) + 5, then divide 6 by 5 and get 6 = 1(5) + 1.
now that the remainder is finally 1, work backwards through the equations to the beginning.
i.e. 1 = 6-5, and since 5 = 11-6, we get that 1 = 6 -(11-6) = 2(6) -11.
and now since 6 = 28 - 2(11) we get 1 = 2(6) -11 = 2(28 - 2(11)) - (11)
= 2(28) -5(11) and since 11 = 39-28 we get 1 = 2(28) - 5(39-28)
= 7(28) - 5(39), i.e. 1 = 7(28) -5(39). now we check it of course since i could be wrong.
140+56 = 196, and 150 +45 = 195. it checks.
the same thing works if you divide the polynomials, provided you use coefficients from a field like the rationals or the reals.
i.e. the key point is that the ring of polynomials over a field forms a euclidean domain.
in order to involve bezout i would suggest that he may well have proved that for polynomials of two variables, say f,g with no common factors, polynomials in x,y, there are two polynomials A,B also in x,y, such that Af+Bg = R(x), where R is a polynomial only in x, called the resultant, but this is usually regarded as due to Euler. Anyway it leads to a proof of bezouts theorem on the number of intersections of the zeroes of f,g, since where f,g, are both zero, then R must be zero, so this let's us find the x coordinates of the poiunts that the curves f=0 and g=0 have in common.
