# Find a basis for Vector space of polynomials

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:$$a_0 + a_1+_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$ 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. Thanks in advance

• Yes, it is quite correct. – Bernard Jan 23 at 15:07

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