How do you solve #-4r ( 3r - 11) ( r - 3) = 0#?

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Answer 1 · 2017-06-18T13:42:06

See a solution process below:

To solve this problem we need to equate each term on the left to #0# and solve for #r#:

Solution 1)

#-4r = 0#

#(-4r)/color(red)(-4) = 0/color(red)(-4)#

#(color(red)(cancel(color(black)(-4)))r)/cancel(color(red)(-4)) = 0/color(red)(-4)#

#r = 0#

Solution 2)

#3r - 11 = 0#

#3r - 11 + color(red)(11) = 0 + color(red)(11)#

#3r - 0 = 11#

#3r = 11#

#3r/color(red)(3) = 11/color(red)(3)#

#color(red)(cancel(color(black)(3)))r/cancel(color(red)(3)) = 11/3#

#r = 11/3#

Solution 3)

#r - 3 = 0#

#r - 3 + color(red)(3) = 0 + color(red)(3)#

#r - 0 = 3#

#r = 3#

The solutions are: #r = 0#; #r = 11/3#; #r = 3#