Because such roots to polynomial equations come in conjugate pairs, we seek the values \( r \) such that \( P(x)=r \) has just one real root \( x \). Considering the shape of a cubic, we are interested in the boundary values \( r \) such that \( P(x)-r \) has a repeated zero. Thus, we write \( P(x)-r=x^{3}+x^{2}-x+(2-r)=(x-p)^{2}(x-q)=x^{3}-(2 p+q) x^{2}+p(p+2 q) x-p^{2} q \). Then \( q=-2 p-1 \) and \( 1=p(p+2 q)=p(-3 p-2) \) so that \( p=1 / 3 \) or \( p=-1 \). It follows that the graph of \( P(x) \) is horizontal at \( x=1 / 3 \) (a maximum) and \( x=-1 \) (a minimum), so the desired values \( r \) are \( r>P(-1)=3 \) and \( r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27 \).
Consider the polynomial \( P(x)=x^{3}+x^{2}-x+2 \). Determine all real numbers \( r \) for which there exists a complex number \( z \) not in the reals such that \( P(z)=r \).