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91Ó°ÊÓ

Find each product. $$\left(3 x y^{2}-4 y\right)\left(3 x y^{2}+4 y\right)$$

Short Answer

Expert verified
The product of the given expression is \(9x^2y^4 - 16y^2\)

Step by step solution

01

Multiply The First Terms

The first terms are \(3 xy^2 \) in each binomial. So, we multiply them to get \( (3xy^2) * (3xy^2) = 9x^2y^4 \)
02

Multiply The Outside Terms

The outside terms are \(3xy^2\) from the first binomial and \(4y\) from the second. We multiply them to get \( (3xy^2) * 4y = 12x y^3 \)
03

Multiply The Inside Terms

The inside terms are \(-4y\) from the first binomial and \(3xy^2\) from the second. We multiply them to get \( -4y*3xy^2 = -12x y^3 \)
04

Multiply The Last Terms

\(-4y\) in the first binomial and \(4y\) in the second are the last terms, we multiply them to get \( -4y*4y = -16y^2 \)
05

Add all obtained results

After multiplication, we add all the obtained results. This cancels out the middle terms as one is positive and the other is negative leading to: \(9x^2y^4 + 12xy^3 - 12xy^3 -16y^2 = 9x^2y^4 - 16y^2\)

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