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Multiply:
$(2xy + 3y^2)$ by $(3y^2 - 2)$
To do:
We have to multiply $(2xy + 3y^2)$ by $(3y^2 - 2)$
Solution:
We know that,
$(a+b)\times(c+d)=a(c+d)+b(c+d)$
Therefore,
$(2xy + 3y^2)\times(3y^2 - 2)=2xy \times (3y^2-2) + 3y^2 \times (3y^2-2)$
$= 2xy \times 3y^2+ 2xy \times (-2) + 3y^2 \times 3y^2 - 3y^2 \times 2$
$= 6xy^{1+2}- 4xy + 9y^{2+2}- 6y^2$
$= 6xy^3 - 4xy + 9y^4- 6y^2$
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