(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
Assuming y' is dy/dx, y = x^4/4 + yx^2
(2xy-4x)+ (8y-16)=2xy-4x+8y-16, which factors as 2(x+4)(y-2)
the difference between 2x2 +4xy-3 and x2-2xy-4 is?
(x - y)2 >= 0 since the left hand side is a square. ie x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy
4xy - 2y(x + 4) = 4xy - 2xy - 8y = 2xy - 8y = 2y(x - 4)
3xy
4
2xy
(2x2+4xy-3)-(x2-2xy-4)Answer: x2+6x+1
(2xy)x=2x2y
Since 2xy is a factor of 2x2y, it is automatically the GCF.