Question 3813
  a(t^2 +1)+ b (t - 1) + c( 2t+2 ) = 0 
 --> a t^2 + (b+2c)t + a-b +2 c = 0
 --> a = 0   (since {t^2, t ,1} is the standard basis of P^2 )
     b+2c = 0  
     a-b+2c = 0  
 --> a = b = c =0

 This shows  t^2 +1, t - 1, 2t+2 are independent and so
 { t^2 +1, t - 1, 2t+2} is a basis of P2 with dim 3.

 Kenny