Multiply (3x-4y)(3x+4y)

Mathematics

1 answer:

mel-nik [20]3 years ago

3 0

You can use (a - b)(a + b) = a² - b²:

$(3x-4y)(3x+4y)=(3x)^2-(4y)^2=3^2x^2-4^2y^2=9x^2-16y^2$

Other method.

Use distributive property: a(b + c) = ab + ac:

$(3x-4y)(3x+4y)=(3x-4y)(3x)+(3x-4y)(4y)\\\\=(3x)(3x)+(-4y)(3x)+(3x)(4y)+(-4y)(4y)\\\\=9x^2\underbrace{-12xy+12xy}_{=0}-16y^2=9x^2-16y^2$

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