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

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Answer 1 · 2016-03-28T04:42:42

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

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

How do you factor the trinomial 3x^2+21xy-54y^23x^2+21xy-54y^2?