MATH.8:5.1 - An even polynomial has more than one solution orbit

For real x, solve P(x)=x^4-5*x^2+4=0. The two transformations are identity and sign reversal r(x)=-x. Since P(-x)=P(x), they preserve the equation.

The known solution 1 gives the orbit {1,-1}. Applying sign reversal again returns to 1, so this orbit is complete. It is not the complete root set: 2 also solves the equation and belongs to the different orbit {2,-2}.

Factorization P(x)=(x^2-1)*(x^2-4) establishes that these two orbits cover all real roots. Symmetry generated each pair; factorization supplied the missing completeness argument.

Now change the equation to Q(x)=x^2+x-2=0. The value 1 remains a root, but Q(-1)=-2. The linear term breaks the sign symmetry. Reusing the old transformation would give a false answer to the changed problem.