thaswasupbreh
thaswasupbreh
16-07-2017
Mathematics
contestada
factors of 3x^2y^2+6x^2+12y^2+24
Respuesta :
Аноним
Аноним
16-07-2017
look for the GCF
GCF = 3
so we have
3 (x^2y^2 + 2x^2 + 4y^2 + 8) Factor by grouping:-
= 3[ x^2(y^2 + 2) + 4(y^2 + 2)]
= 3(x^2 + 4)(y^2 + 2)
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