How do you solve #x( x - 2) ( x + 1) = 0#?

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Answer 1 · 2016-11-13T22:01:42

#x=0"; "x=+2"; "x=-1#

#color(red)("This question has a trap")#

If you were to divide both sides by #(x+1)# you would end up with

#x(x-2)=0# which will give 2 solutions

#color(green)("There is a problem with the above approach")#

You would miss one of the solution

The three possible scenarios are

#x=0#
#(x-2)=0#
#(x+1)=0#

So as solutions we have:

#x=0"; "x=+2"; "x=-1#