Let (x+a)(x+b)=x2+px+q and (x+a)(x+c)=x2+mx+n, where x+b and x+c are also factors, which combine with x+a to produce the two assumedly different quadratics (b and c are unknown), then:
p=a+b and q=ab, m=a+c, n=ac. From this b=q/a and c=n/a. Now we can write:
p=a+q/a and m=a+n/a, so p-m=(q-n)/a, or a(p-m)=q-n. These equations contain no references to the unknowns b and c, and they relate the unknowns m, n, p, q in terms of a.
We can also create two quadratics in a:
a2-ap+q=0 and a2-am+n=0 from which a=(p±√(p2-4q)/2=(m±√(m2-4n)/2, which tells us that p2≥4q and m2≥4n.
Note that since x+a can be factors of many different polynomials, we cannot relate m and n to p and q.
For example: (x+2)(x+3)=x2+5x+6, (p=5, q=6) and (x+2)(x+7)=x2+9x+14 (m=9, n=14). But x+3 and x+7 are arbitrary factors and p and q are independent of m and n, just as 3 and 7 are independent of one another. The only fixed relationships are between pairs p and q, and m and n, because they belong to the same quadratic.