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#