2+2y+x+xy=(x+2)(y+1)
I understand you want [(2/x) + (1/y)] as a single fraction. The common denominator is (xy): 2/x = (2y)/(xy); and 1/y = x/(xy), so the answer is (2y + x) / (xy)
if you take a factor of (Y) you end up with Y(X+5Z-2).
2+2y+x+xy
(xy - 7)(x^2y^2 + 7xy + 49)
(x + 2)(y - 3)
xy(x - 2y)(x + 2y)
0
(y - 3)(x + 2y)
(1 - x4y4) = (1 + x2y2)*(1 - x2y2) = (1 + x2y2)*(1 + xy)*(1 - xy)
(xy + 7)(x^2y^2 - 7xy + 49)
If that's xy + x^2y, it factors to xy(x + 1) If it's something else, please re-submit your question with any plus signs written out.
Factor out a 2; so it is 2(x+2y).
First it would be easiest to factor out the 2. 2x2-8y2 2(x2-4y2) Then you factor the part in the parenthesis as follows 2(x-2y)(x-2y) which can be simplified as 2(x-2y)2
(x - 2y^2)(x + 2y^2)(x^2 + 4y^4)
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
xy + x = x(y + 1)