# (i) show that p1, p2 and p3 form a basis for R[x]=<2. (ii) Find g belongs to R[x]=<2 so that g(-1) = 0, g(0) = -12, g(1) = 1. p1(x) := ((1/2)^x^2) - ((1/2)^x), p2(x) := -x^2 +1, p3(x) := ((1/2)^x^2) + (1/2)^2 I changed `P_3` to `(1/2)X^2+(1/2)X`` `. I guess there is a typo in the question.

(i) P1, P2, P3 are 3 polynomials of the 3 dimensional vector space `RR_2[X].` It is a basis iff they are linearly independant.

Solve `aP_1+bP_2+cP_3=0`

`a(1/2) x^2-a(1/2)x+b(-x^2)+b+c(1/2)x^2+c(1/2)x=0`

Sort the coefficients.

`(a/2-b+c/2)x^2+((-a/2+c/2)x+b=0`

A polynomial that is identically 0  has all its coefficients null.

Therefore `a/2-b+c/2=0 ` and `(-a^2+c/2)=0 ` and b=0

Therefore b=0 and a+c=0 and a-c=0

Therefore b=0 and a+c=0 and c=a

Therefore b=0 a=0 c=0, the 3 polynomials are linearly independant.

They are a basis of `RR_2[X]`

(ii) P_1, P_2, P_3 generates `RR_2[X] ` therefore there exists (a,b,c) such that `g=aP_1+bP_2+cP_3.`

`g(-1)=0=aP_1(-1)+bP_2(-1)+cP_3(-1)`

`g(-1)=0=a*1+0+0`
therefore a=0

`g(0)=-12=aP_1(0)+bP_2(0)+cP_3(0)`

`g(0)=-12=0+b+0`

therefore b=-12

`g(1)=1=aP_1(1)+bP_2(1)+cP_3(1)`

`g(1)=1=0+0+c*1`

Therefore c=1

Therefore `g=-12P_2+p_3`

`g=12X^2-12+(1/2)X^2+1/2 X`