How do you factor #x^3 - 2x^2 - x + 2 #?

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Answer 1 · 2015-05-22T13:49:42

We can factor out #x# from all the elements that contain it:

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

And then factor the quadratic inside the parenthesis, which, using Bhaskara, will give us the following roots:

#(2+-sqrtcancel((4-4(1)(1))))/2#
#2/2=1#

#x_1=x_2=1#

Thus, the factors are both #(x-1)#

Finally: #x((x-1)(x-1))+2#

Answer 2 · 2015-05-22T14:12:22

Factor by grouping:

#x^3-2x^2-x+2#

#=(x^3-2x^2)+(-x+2)# (Notice the inserted addition before the second pair.)

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

#=(x^2-1)(x-2)# Now the first binomial is a difference of squares, so

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

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