What is the solution to the following system? 2x - y - 3z = 1
4x + 3y + 2z = -4
-3x + 2y + 5z = -3

(2, - 6, 3)

Step-by-step explanation:

2x - y - 3z = 1 ⇒ y = 2x - 3z - 1 (1)

4x + 3y + 2z = - 4 (2)

- 3x + 2y + 5z = - 3 (3)

(1) ---> (2) and (3)

4x + 3(2x - 3z - 1) + 2z = - 4 ⇔ 10x - 7z = - 1 ... (4)

- 3x + 2(2x - 3z - 1) + 5z = - 3 ⇔ x - z = - 1 ⇒ x = z - 1 ---> (4)

10(z - 1) - 7z = - 1 ⇒ z = 3

x = 2

2(2) - y - 3(3) = 1 ⇒ y = - 6

(2, - 6, 3)

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option