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,

$(2xy + 3y^2)\times(3y^2 - 2)=2xy \times (3y^2-2) + 3y^2 \times (3y^2-2)$

$= 2xy \times 3y^2+ 2xy \times (-2) + 3y^2 \times 3y^2 - 3y^2 \times 2$

$= 6xy^{1+2}- 4xy + 9y^{2+2}- 6y^2$

$= 6xy^3 - 4xy + 9y^4- 6y^2$

Tutorialspoint

Updated on: 10-Oct-2022

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