What is the solution set for #-x^3-2x^2=-x-2#?

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Answer 1 · 2015-08-22T04:53:55

The solution set is #x={-2,-1,1}#

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

Multiply the equation times #-1#.

#x^3+2x^2=x+2#

Move all terms to the left side by subtracting #x+2# from both sides.

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

Group the terms into two binomials.

#(x^3+2x^2)-(x+2)#

Factor out #x^2# from the first group.

#x^2(x+2)-(x+2)#

Factor out #(x+2)#.

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

#(x^2-1)# can be factored as the difference of squares.

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

The complete factorization of #-x^3-2x^2=-x-2# is #(x+2)(x+1)(x-1)=0#.

Since the factors are equal to zero, we must set each one equal to zero, and solve for #x#.

#x+2=0#
#x=-2#

#x+1=0#
#x=-1#

#x-1=0#
#x=1#

The solution set is #x={-2,-1,1}#