SOLUTION: Factor completely: 36(x^4)y + 12(x^3)y^2 - 15(x^2)(y^3) Thank you D. Scarfone

Question 2646: Factor completely: 36(x^4)y + 12(x^3)y^2 - 15(x^2)(y^3)
Thank you
D. Scarfone
Answer by kiru_khandelwal(79) (Show Source): You can put this solution on YOUR website!
36(x^4)y + 12(x^3)y^2 - 15(x^2)(y^3)
=> taking out x^2 as common
=> 3x^2(12(x^2)y + 4xy^2 - 5y^3)
Taking out y as common factor
=> 3(x^2)y(12x^2 + 4xy - 5y^2) = 0
=> either x^2 = 0 => x = 0
or y = 0
or 12x^2 + 4xy - 5y^2 = 0
=> 12x^2 -6xy + 10xy - 5y^2 = 0
=> 6x(2x-y) + 5y(2x-y)=0
=> (6x+5y)(2x-y)=0
=> According to the zerp product rule
either 6x + 5y =0........equation(1)
or 2x -y =0 => y = 2x...........equatin(2)
substituting the value of y from equation(2) to equation(1)
=> 6x + 5y =0
=> 6x + 5(2x) = 0
=> 16x = 0 => x = 0
and since y = 2x = 0
So the only solution is x=0 and y=0
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