Conversion of $F(x) = {1 \choose 0} (2t^2 - 3t + 1) + {3 \choose 1} (4t - 4t^2) + {1 \choose 2} (2t^2 - t)$

Question

I have a textbook with a calculation step that is pretty unclear to me:

$$F(x) = {1 \choose 0} (2t^2 - 3t + 1) + {3 \choose 1} (4t - 4t^2) + {1 \choose 2} (2t^2 - t)$$

$$= {-8 \choose 0} t^2 + {8 \choose 2} t + {1 \choose 0}$$

I guess there is some very simple math involved that I just don't know at all :.-(

Answer 1

Coordinate-wise, you have

$$F_1(t) = 1(2t^2-3t+1)+3(4t-4t^2)+1(2t^2-t) = 2t^2-3t+1-12t^2+12t+2t^2-t$$

Group like terms:

$$F_1(t) = -8t^2+8t+1.$$

Perform a similar association with the second "row".