Firstly, sorry for poor definition of the following example, I translated it from my native language,hope you can understand
So i need help with this:
Mapping is given $\langle , \rangle:R_{2}[x] \times R_{2}[x] \rightarrow R$ by: $$\langle a+bx+cx^2, \alpha+\beta x+\gamma x^2 \rangle = a\alpha+2b\beta+2c\gamma-a\beta-b\alpha-b\gamma-c\beta$$ Check if scalar product of space $R_{2}[x]$ is given by mapping $\langle , \rangle$
So as far as I know to check if this is a scalar product we need to check next:
1.$\langle x,x \rangle\geq 0 , \forall x\in E$, E is real vector space
$\langle x,x \rangle = 0\Leftrightarrow x=0$
2.$\langle x,y \rangle = \langle y,x \rangle$
3.$\langle \lambda x,y \rangle = \lambda\langle x,y \rangle$, $\lambda \in R$
4.$\langle x+y,z \rangle = \langle x,z \rangle + \langle y,z \rangle$
I know how to do this for simple examples like when $x$ is given only with $x=\lbrace \alpha_{1},\alpha_{2},\alpha_{3}\rbrace$,but for something like this above i got confused
Your usage of x and y in the definiton is confusing because x also appears in the polynomial. Your x and y are now polynomials of degree 2, so your real vectorspace now is the vectorspace of polynomials with degree 2. For a better understanding write the definitons with polynomials $p$, $q$ and $p=a+bx+cx^2$, $q=\alpha + \beta x + \gamma x^2$.
Then you have got to Show for example $(p, p) \geq 0$ for All polynomials $p \in R_2 [x] $.
Example: Let $p \in R_2[x]$, so $p=a +bx + cx^2$ for some $a,b, c \in R$. Then $0 = (p, p) =a^2 + 2b^2 + 2c^2 - 2ab - 2bc =(a-b)^2 + (b-c)^2 + c^2$, which is equivalent to $(a-b) ^2 =0$, $(b-c)^2=0$ and $c^2=0$ (since for any real number $r$ it holds $r^2 \geq 0$). Now you can go on with equivalences until you get to $p=0$. But notice that $p$ is a polynomial, so all coefficients $a, b, c$ have to be 0.