How to solve #x^3-3x-2=0# ?

2 Answers
Ananda Dasgupta · MathFact-orials.blogspot.com
Jun 3, 2018

The roots are #-1,-1,2#

Explanation:

It is easy to see by inspection that #x = -1# satisfies the equation :

#(-1)^3-3times(-1)-2 = -1+3-2=0#

To find the other roots let us rewrite #x^3-3x-2# keeping in mind that #x+1# is a factor:

#x^3-3x-2 = x^3+x^2-x^2-x-2x-2#
#qquadqquad = x^2(x+1)-x(x+1)-2(x+1)#
#qquadqquad = (x+1)(x^2-x-2)#
#qquadqquad = (x+1)(x^2+x-2x-2)#
#qquadqquad = (x+1){x(x+1)-2(x+1)}#
#qquadqquad = (x+1)^2(x-2)#

Thus, our equation becomes

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

which obviously has roots #-1,-1,2#

We can also see it in the graph:

graph{x^3-3x-2}

Cem Sentin
Jun 4, 2018

#x_1=x_2=-1# and #x_3=2#

Explanation:

#x^3-3x-2=0#

#x^3+1-(3x+3)=0#

#(x+1)(x^2-x+1)-3(x+1)=0#

#(x+1)(x^2-x+1-3)=0#

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

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

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

Thus #x_1=x_2=-1# and #x_3=2#

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