Let $ p(x)=\alpha x^{2}+\beta x+\gamma $ be a polynomial Let $ p(x)=\alpha x^{2}+\beta x+\gamma $  be a polynomial , where $ \alpha, \ \beta , \gamma \in \mathbb{R} $. Fix  $ x_{0} \in \mathbb{R}$. Let $ \ S=\{(a,b,c): p(x)=a(x-x_{0})^{2}+b(x-x_{0})+c \}$. Then cardinality of $ S $ is
(a) $ 0 $, 
(b) $ 1 $,
(c) $ \infty $
My approach:   I thought the option $ (c) $ is true. But not sure . Any help ?

We can easily find $a,b,c$ such that $p(x)=a(x-x_{0})^{2}+b(x-x_{0})+c$ for any $x\in\Bbb{R}$, so $S$ is not empty. 
To show that $\{a,b,c\}$ is the only element of $S$, note that a polynomial is the zero polynomial if and only if all it's coefficients are equal to $0$.
Therefore the cardinality of $S$ is $1$.
