# Find a basis for $p \in P_2(\mathbb R)$ with $p(7) = 0$?

Find a basis for the subset: $$S = \{\;p \in P_2(\mathbb R)\;\; |\;\; p(7) = 0\; \}$$

I'm not sure how to approach this question.

$$p(7) = a_0 + 7a_1 + 49a_2 = 0$$ $$a_0 = -7a_1 - 49a_2$$

$$a_1(-7 + x) + a_2(-49 + x^2) = 0$$

Is this even in the right direction? I don't know what to do from here, nor how to proceed by manipulating the standard basis $\{1,x,x^2\}$

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Hint: A polynomial $p$ has $p(7) = 0$ if and only if $(x - 7) | p(x)$. That is, if we can write

$$p(x) = (x - 7)(ax + b)$$

since $p$ has degree $2$. Now $a$ and $b$ give two free choices.

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I tried that approach too, but I'm not sure what to do next...With a vector subspace, I simply rearrange one of the terms to be dependent on the other terms, and then find a basis. But I'm really confused here. – tgun926 Nov 4 '13 at 4:23
@tgun926 Take $a = 1, b = 0$ to get $x^2 - 7x$, and vice-versa to get $x - 7$. Prove that these are independent / span. – user61527 Nov 4 '13 at 4:28