Good! A more fundamental approach, using the definition of "linearly independent"- suppose a(1)+ b(-1+ x)+ c(1+ x^2)= 0 for all x. Taking x= 0, a(1)+ b(-1)+ c(1)= a- b+ c= 0. Taking x= 1, a(1)+ b(0)+ c(2)= a+ 2c= 0. Taking x= -1, a(1)+ b(-2)+ c(2)= a- 2b- 2c= 0. Subtracting the second equation from the first eliminates a: - b- c= 0. Subtracting the third equation from the first b+ 3c= 0. Adding those two equations, 2c= 0 so c= 0 and then b= a= 0.

Do you have the theorem that "any linearly independent set containing a number of vectors equal to the dimension of the space is a basis" and that the dimension of P2 is 2? If so, then yes, showing that those three vectors are independent show this is a basis. If not, then you should show that for any a, b, c, there exist , , such that . That's, fairly easy- just solve for , , and .

No, your "vector" is a+ b+ cx^2 for(b) I know this is incorrect, but I got:

p(x) = a(1,0,0) + b(-1,1,0) + c(1,0,1)

a=b=c=1 (from (a))

Thereforeanya, b, c- you can't just make a= b= c= 1. Your coefficients must be in terms of a, b, and c- just what I said in my last sentence above! Find , , such that , forallx.

One way to do this is to equate like coefficients. Multiplying out the left side, for all x so we must have , and . Putting those into the first equation, so :

.

Use the answer to (b) to write out and then . The "coordinates" of each are the number , , and(c) No idea?

Please help!