A bag contains 6 red marbles, 2 blue marbles, and 1 green marble. What is the
probability of choosing 2 marbles that are not green when 2 marbles are drawn
from the bag? (no replacement before choosing second marble)

We have 1 green and 8 non-green.  

We get a probability of 8/9 of a non-green for the first marble and a probably of 7/8 for the second being non green, so

p = (8/9)*(7/8) = 7/9

7/9

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form