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first let me start by apologizing if this question has been asked before, but i've never studied math in english and am lacking the language to formulate a proper search string, and now onto the question:

let $V$ be a vector space of the real polynomials where the following is true: $f(-1)=f(1)=0$

the inner product is defined as: $\langle f, g \rangle = \displaystyle\int_{-1}^{1} f(x)g(x)dx$

i need to find a orthonormal base for $V$. i can't seem to figure out how to use f(-1)=f(1)=0 some one told me that it is supposed to help me pick a base, but i don't see how, any sort of hint/solution would be greatly appreciated

note: i found this question which seems to be almost identical but without $f(-1)=f(1)=0$

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    The vector space of these polynomials is the set of polynomials of the form $((x^2-1)P(x)$ where $P(x)$ is any polynomial. – 2017-02-25
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    @JeanMarie im sorry, i don't know what that means – 2017-02-25
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    What @JeanMarie meant is the following: Let $q(x)$ be a real polynomial such that $q(-1)= q(1) =0$. Then $(x+1)/q(x)$ and $(x-1)/q(x)$, in particular $(x^2 -1)/ q(x)$ (why?). Thus, one can write $q(x) = (x^2-1)p(x)$, where $p(x)$ is another real polynomial. – 2017-02-25
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    @GiovanniDeGaetano unfortunately, i still don't understand how this helps me pick a base. should i attempt to pick things that are divisible by $q(x)$ ? – 2017-02-25
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    Do you know a basis for the whole space of real polynomials? Then you can express $p(x)$ in terms of this basis and write $q(x) = (x^2-1)p(x)$. Does it make sense? – 2017-02-25
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    @GiovanniDeGaetano if my base is {$1$,$x$,$x^{2}$,$x^{3}$} do i just multiply them by $(x^2 - 1)$ ? that cant be right.. do i try to represent them as $(x^2 - 1)$ times something = my base members? (i don't actaully know much calculus, this is a particularly bothersome question in an algebra course) – 2017-02-25
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    @Nullman that's right, the basis is $(x^2-1).1, (x^2-1).x, (x^2-1).x^2, ...$ – 2017-02-25
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    @JeanMarie i think i'm misunderstanding something basic here. lets say i want to preform $⟨f,f⟩$ on $x^2 - 1$ dont i just end up with a more complex expression then i would have if i did $⟨1,1⟩$ ? – 2017-02-25

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