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I would like to find a basis for the vector space of Polynomials of degree 3 or less over the reals satisfying the following 2 properties: $$p(1)=0$$ $$p(x)=p(-x)$$

I started with a generic polynomial in the vector space: $$a_0 + a_1x+a_2x^2+a_3x^3$$ and tried to make it fit both conditions:$$a_0 + a_1+a_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$ the second equations becomes $$a_1x+a_3x^3=0$$ thus $a_1$ and $a_3$ must be constantly equal to 0. Plugging back into the first equation we get $$a_0 = -a_2$$ thus $p=a_0 -a_0x^2$. Then $1-x^2$ would be a basis? Is my method correct? If not, how would one solve this type of problem.

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    $\begingroup$ Yes, it is quite correct. $\endgroup$
    – Bernard
    Jan 23, 2019 at 15:07

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Yes, that is correct. The vector space has dimension $1$ and $\{1-x^2\}$ is a basis.

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