One Solution
(x^24x+7)*(y^2+2y+6) = 15
(x^24x+4 + 3)*(y^2+2y+1 + 5) = 15
((x2)^2 + 3)*((y+1)^2 + 5) = 15
if (x,y) = (2,1) then:
3 * 5 = 15
So one solution is (x,y) = (2,1) and requested product x*y = 2
One could set the pair of quadratics to all the combination of factors that multiply to 15:
(x^24x+7) , (y^2+2y+6) = 3 , 5 has already been done above
But there is still:
A , B
1 , 15
5 , 3
15 , 1
1 , 15
3 , 5
5 , 3
15 , 1
Each would produce 2 roots for f(x) = A and 2 roots for g(y) = B, potentially 4 values for x*y.
Will do for the first example of 1,15
x^24x+7 = 1 y^2+2y+6 = 15
x^24x+6 = 0 y^2+2y9 = 0
x roots: 2 ± √2 i y roots: 1 ± √10
There are 4 different complex values for an x root times a y root; but the arithmetic average is still 2.
I suspect this holds true for the other examples but I have not checked.

Posted by Larry
on 20240508 13:02:30 