thaswasupbreh
thaswasupbreh thaswasupbreh
  • 16-07-2017
  • Mathematics
contestada

factors of 3x^2y^2+6x^2+12y^2+24

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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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