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How do you factor the trinomial #3x^2+21xy-54y^2#?

1 Answer

Konstantinos Michailidis

Mar 28, 2016

We have that

#3x^2+21xy-54y^2=3*(x^2+7xy-18y^2)=3*(x^2+9xy-2xy-18y^2)= 3*(x^2-2xy+9xy-18y^2)=3*(x(x-2y)+9y(x-2y))= 3*(x-2y)*(x+9y)#

Finally

#3x^2+21xy-54y^2=3*(x-2y)*(x+9y)#

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