Factor x^3 + x^2 - x - 1 completely.

Step-by-step solution

1. 1

   Group the four terms in pairs. A four-term cubic with no common factor is the classic grouping candidate:

   $\left(x^3+x^2\right)+(-x-1)$

2. 2

   Factor each group so a common binomial appears. Take $x^2$ from the first pair and $-1$ from the second:

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

   Pulling out $-1$ rather than $+1$ is the key move: $-x-1=-1(x+1)$.

3. 3

   Extract the common binomial.

   $x^3+x^2-x-1=(x+1)\left(x^2-1\right)$

4. 4

   Do not stop — factor the difference of squares. $x^2-1=(x-1)(x+1)$, which produces a repeated factor:

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

5. 5

   Verify by expanding and by roots. $(x+1)^2(x-1)=(x^2+2x+1)(x-1)=x^3+x^2-x-1$ $\checkmark$. The roots are $x=-1$ (double) and $x=1$; substituting $x=-1$ into the original gives $-1+1+1-1=0$ $\checkmark$.