How do you find the zeroes of #p(x)= x^3+x^2-x-1#?

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Answer 1 · 2015-05-11T23:19:53

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

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

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

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

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

This is zero for #x = -1# (twice) and #x = 1#.

The process I used here is called factoring by grouping.