Question
Multiply:
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
✓
Answer
To multiply, we will use distributive law as follows:
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
$ =2 x y\left(3 y^2-2\right)+3 y^2\left(3 y^2-2\right) $
$ =6 x y^3-4 x y+9 y^4-6 y^2 $
$ =9 y^4+6 x y^3-6 y^2-4 x y $
Thus, the answer is $ 9 y^4+6 x y^3-6 y^2-4 x y $
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
$ =2 x y\left(3 y^2-2\right)+3 y^2\left(3 y^2-2\right) $
$ =6 x y^3-4 x y+9 y^4-6 y^2 $
$ =9 y^4+6 x y^3-6 y^2-4 x y $
Thus, the answer is $ 9 y^4+6 x y^3-6 y^2-4 x y $
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