Math (precalculus)

I have a question about the symmetry of graphs, but maybe it's more of a simple factoring question...

Why is f(x)=x+(1/x) odd, while h(x)=x-x^2 is neither even nor odd?

I understand that f(-x)=-x-1/x=-(x+1/x)=-f(x) is odd because f(x)=f(-x).

Then for h(-x)=-x-x^2...why can't you factor out the negative in this question like the previous one? But my textbook leaves it like this and says it's neither even nor odd...

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asked by Lucy
  1. h(x) = x - x^2
    h(-x) = -x -x^2
    but x is not -x so that part is odd
    while -x^2 is -x^2 so that part is even

    for odd I needed
    h(-x) = -x + x^2 = -(x-x^2)
    for even I needed
    h(-x) = x - x^2

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    posted by Damon

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