How do you factor #x³+x²-x-1#?

2 Answers
Nov 13, 2015

Answer:

The complete factorization is #(x+1)^2(x−1)# or #(x+1)(x+1)(x−1)#.

Explanation:

I used synthetic division to solve this. (Do you need further explanation?)

Nov 13, 2015

Answer:

You can factor by grouping to find:

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

Then use the difference of squares identity to find:

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

Explanation:

First factor by grouping:

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

Then notice that #x^2-1 = x^2-1^2# is a difference of squares, so we can use the difference of squares identity [ #a^2-b^2 = (a-b)(a+b)# ] to find:

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

Alternatively, notice that the sum of the coefficients (#1+1-1-1#) is #0#, so #x=1# is a zero of this cubic polynomial and #(x-1)# is a factor.

Divide #x^3+x^2-x-1# by #(x-1)# to get #x^2+2x+1# :

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Then recognise that #x^2+2x+1 = (x+1)^2# is a perfect square trinomial. One little trick to spot this one is that #11^2 = 121#, the #1,2,1# matching the coefficients of the quadratic and #1,1# matching the coefficients of the linear factor.