How do you multiply $(3x-4) (4x+3)$ ?

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Answer 1 · 2015-06-26T23:04:29

$(3x-4)(4x+3)$

$= 3x(4x+3)-4(4x+3)$

$= 12x^2+9x-16x-12$

$= 12x^2-7x-12$

This is like the reverse of factoring by grouping, using the distributive property of multiplication...

$(3x-4)(4x+3)$

$= 3x(4x+3)-4(4x+3)$

$= (3x*4x)+(3x*3)-(4*4x)-(4*3)$

$= 12x^2+9x-16x-12$

$= 12x^2+(9-16)x-12$

$= 12x^2-7x-12$

How do you multiply (3x-4) (4x+3)(3x-4) (4x+3)?