How do you solve #(2x+8)(3x-6)=0# using any method?

1 Answer
John C.
May 4, 2017

#x=-4 or x=2#

Explanation:

Expand the equation from the brackets

#(2x+8)(3x-6)=0#
#(color(red)(2x)#+8)(#color(red)(3x)-6)# Multiply the highest power of #x# first
#(color(red)(2x)#+8)(3x#color(red)(-6)#) Next multiply by the next highest power

(2x#color(blue)(+8#)(#color(blue)(3x)#-6) Multiply by the highest power
(2x#color(blue)(+8#)(3x#color(blue)(-6)#) Multiply by next highest power.

Therefore,
#6x^2-12x+24x-48=0#
#6x^2+12x-48#=0 ------------------Divide the equation by 6
#x^2+2x-8=0# --------------------------Simplify the equation
#(x+4)(x-2)=0#
#x+4=0 or x-2=0#
#x=-4 or x=2#

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