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Do all polynomials $ax^3 + bx^2 + cx + d$ with a root at $x=1$ form a vector space? Do the coefficients $(a,b,c,d)$ form a vector space?

My reasoning: Since $x=1$ is a root, we can't have $(a,b,c,d)$ all zeros. The space doesn't include the zero-matrix. Hence, it's not a vector space

Suppose, $f(x)$ and $g(x)$ = polynomials of form $ax^3 + bx^2 + cx + d$ with root at $x=1$ $h(x) = f(x) + g(x)$ Since $x=1$ is a root, $f(1) = 0$ and $g(1) = 0$. $h(1) = f(1) + g(1) = 0$ ($h(x)$ also has a root at $x = 1$)

I'm not sure about the zero matrix in this case. What is considered as "zero matrix" in polynomials?

Thanks for the help in advance.

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No. If you mean by "zero matrix" the zero vector, I don't understand why you think the set of polynomials with root at $1$ does not contain the zero vector. The zero vector $(0,0,0,0)$ corresponds to the zero polynomial and is certainly also $0$ at $1$. – Rudy the Reindeer Sep 19 '12 at 12:04

Hint: For a polynomial $ax^3+bx^2+cx+d$ having a root at $x=1$ is equivalent to have $a+b+c+d=0$.

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The zero vector in this case is $(0,0,0,0)$. This corresponds to the polynomials $p(x) = 0$. You have $p(1) = 0$ hence $p$ is in the set of all polynomials that are zero at $1$. So your set does contain the zero vector.

You have already shown that the set is closed with respect to addition and now we know that it also contains the zero vector. The only thing that remains to be verified is that it is also closed with respect to scalar multiplication.

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