Here's an alternative approach.
By the Rational Root Theorem we know
\(\displaystyle P(x) \ = \ x^4 \ + \ 2x^3 \ - \ x \ - \ 1\)
Does not have any rational roots. This means P(x) cannot be reduced to anything that has a rational linear factor. There remains only one possibility to test , can it be reduced to the product of two quadratic factors with rational coefficients?
\(\displaystyle x^4 \ + \ 2x^3 \ - \ x \ - \ 1 \ = \ (x^2 \ + \ ax \ + \ 1 )(x^2 \ + \ bx \ - \ 1) \ =\)
\(\displaystyle x^4 \ + \ (b \ + \ a)x^3 \ + \ abx^2 \ + \ (b \ - \ a)x \ - \ 1\)
Itis clear that either a or b or both must be 0 otherwise we get a x^2 term which we don't want.
If both a , b = 0 will lead to
\(\displaystyle (x^2 \ + \ 1)(x^2 \ - \ 1)\)
contradicting RRT. WLOG let a = 0 , then we are forced to write
\(\displaystyle (x^2 \ + \ 1)(x^2 \ + \ bx \ - \ 1)\)
\(\displaystyle x^4 \ + \ bx^3 \ + \ bx \ - \ 1\)
Comparing to P(x) we get b = 2 and -1 , impossible.
Note: I'm not comfortable with this 'analysis'