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(1) y2-2zy-2y-2z-3 = y2 -2y (z+1) - (2z+3)
Delta = [-(z+1)]2 - (- (2z+3)) = z2+2z+1+2z+3= z2+2z+4 = (z+2)2
(If used the reduced form as b=-2(z+1) is even, b'=-(z+1) and delta=b'2-ac)
y=(z+1) ± (z+2) so y = 2z+3 or y =-1
So the (1) can be written (y-2z-3)(y+1)
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