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$$P(x) = \frac{x^2-1}{x+1}$$

Hi, If I factor $x^2 - 1$, I get $(x+1)(x-1)$, which can be divided by $(x+1)$. But if I leave the expression as it is and give $x$ a value of $-1$, this won’t be a polynomial.

So is this expression a polynomial?

user26857
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    Technically, no, as it is ill defined at $x=-1$. – lulu Jan 11 '22 at 19:58
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    It is not a polynomial, but a rational function. A rational function is a ratio of polynomials (for denominator not identically zero): https://en.wikipedia.org/wiki/Rational_function – Golden_Ratio Jan 11 '22 at 20:02
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    You are right thinking that $x=-1$ is a problematic value. Polynomials are defined on $\mathbb{R},$ while $P(x)$ is not. – user376343 Jan 11 '22 at 20:03
  • @user26857: I suppose $x^2 + 2x + 0\cdot\sin x$ is a polynomial function that isn't a (formal, in the algebraic sense) polynomial? (I haven't tried looking up the formal definitions yet.) – Dave L. Renfro Jan 11 '22 at 20:23

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From the definition of your function it is obvious it is not defined for $x=-1$, hence it is not a polynomial. A polynomial must be defined for every real argument.

Artur Wiadrowski
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