How do you factor completely #x^3+x^2-x-1#?

1 Answer
Nov 29, 2015

Answer:

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

Explanation:

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

First notice that the sum of the coefficients is #0# so #f(1) = 1+1-1-1 = 0# and #x-1# is a factor of #f(x)#.

Divide by #(x-1)#...

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Then factor the remaining quadratic factor - recognisable as a perfect square trinomial.

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

A little trick to spot that #x^2+2x+1# is #(x+1)^2# is given by the pattern of the coefficients: #1, 2, 1# is the same pattern of digits as #121 = 11^2#.