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What will be the basis for the space of cubic polynomials $p$ such that $p(3) = 0$?

I know that: A natural basis for the vector space of cubic polynomials is $p(3)$ is $\langle 1, x, x^2, x^3 \rangle$.

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    Look for a polynomial of degree $1$, one of a degree $2$ and one of degree $3$ such that $p(3)=0$.2011-09-23
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    But to find p(3) = 0 , how will I apply the natural basis for vector space of cubic polynomials to the question?2011-09-23
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    Cheap route: The basis for the cubic polynomials such that $P(0)=0$ is just those without a $x^0$ term: $\{x,x^2,x^3\}$. So one can argue the basis for cubic polynomials such that $P(3)=0$ will be $\{(x-3),(x-3)^2,(x-3)^3\}$.2011-09-23
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    To Narayanan, I meant "a" basis, thank you for clarifying that.2011-09-23
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    Ah, I also said "the." Oh well, brain leak.2011-09-23
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    I have some nitpicks wrt this question. (1.) You cannot ask for *the* basis since a vector space might have many (even uncountably many) bases. (2.) A basis is a set, so perhaps you can use braces to represent it, like $\{ 1, x, x^2, x^3 \}$, rather than $\langle \rangle$. (3.) You presumably mean the space of polynomials of degree *at most* $3$. The space of polynomials of degree exactly $3$ is not a linear space. (I am not sure what the term "cubic" conventionally means.) My apologies if this feels like empty nitpicking, but I think it's good to be correct :-).2011-09-23
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    @Blake You really need to be speaking of the vector space of polynomials of degree *less than or equal to* 3.2011-09-23
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    Take a basis of the space of degree at most $2$ polynomials, and multiply it by $x-3$.2011-09-23

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