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Some days ago, I've made this question and I guess I've finally found an answer to this question:

  • (a) Is it possible to find a polynomial, apart from the constant $0$ itself, which is identically equal to $0$ (i.e. a polynomial $P(t)$ with some nonzero coefficient such that $P(c)=0$ for each number $c$)?

Which is mentioned in that post, I tried to proceed in the following way: I decomposed the polynomial in 2 kinds of structures - due to their similar behaviors inside the polynomial - for example:

$$\color{red}{a_nt^n}+\color{red}{a_{n-1}x^{n-1}}+\color{red}{...}+\color{red}{a_1t}+\color{green}{a_0}$$

The red structures are the $t$'s (1) with a coefficient and the green structure is a number alone, for the red structures, it's impossible to have $P(c)=0$ for every number $c$, with nonzero coefficients. Then I eliminated the reds and started to think about the green alone and I perceived that the only number which would allow me to find this polynomial was zero, thus making it impossible to find the requested polynomial.

Is this valid?

(1) - I have no idea on how to call the $t$'s in the polynomial, the book I'm reading provided me with some terms like: Leading coefficient, coefficients, leading term, constant term/constant coefficient, linear coefficient and linear term, but it mentions no name for the $t$'s, can you provide me a name for it?

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    Are you perhaps trying to induct on the number of non-zero coefficients?2012-10-14
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    Individual $t$ is the indeterminate of the polynomial (and I suppose the $x$ in the second term should be $t$ as well), the powers $t^i$ are monomials. Instead of "structure" most people say "term". Also your argument is fundamentally flawed, because say a sum of two part (red and green) is always zero does not imply each of the parts is always zero. Consider $\color{red}{-x}+\color{green}{x}$.2012-10-14

4 Answers 4