How do you solve by factoring and using the principle of zero products #36r^2=16#?

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Answer 1 · 2017-01-03T22:06:32

#r=2/3#
#r=-2/3#

#36r^2=16#

Make one side equal to zero by subtracting 16 from each side:
#36r^2-16=0#

First, I would factor out a #4# so that I'm working with smaller numbers:
#4(9r^2-4)=0#

Use the property that: #a^2-b^2=(a+b)(a-b)#

#4[(3r)^2-(2)^2]=0#

#4(3r-2)(3r+2)=0#

Set each factor (apart from constants) equal to zero individually.
#3r-2=0#
#3r=2#
#r=2/3#

#3r+2=0#
#3r=-2#
#r=-2/3#