Solve the system. -a+4b+2c=-8 3a+b-4c=9 b=-1 a. (0,1,1) b. (-1,-1,2) c. (1,1,-2) d. (2,-1,-1)

Option D is right,.

Step-by-step explanation:

-a+4b+2c=-8   ... i

3a+b-4c=9      ... ii

b=-1                 ... iii

Substitute the value of b, in i and ii

-a-4+2c =-8 or -a+2c = -4

and 3a-1-4c = 9 or 3a -4c =8

Now we have two equations in two variables

-a+2c =-4 and 3a-4c =8

a = 2c+4:  substitute this in the other equatin.

3(2c+4)-4c =8 Or 2c +12 =8

2c =-4 or c = -2

Substitute in -a+2c =12

-a-2 = -4

a = -2+4 =2

a=2, b=-1 and c =-1 is the solution.

- 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