Question

plz ans it fast ..........​

Answer 1

Answer:

-3,4 ...............................

Answer 2

$\red{Given\::-}$

• x² + 1/6 x - 2

$\orange{To\:Find\::-}$

• Zeroes of the given polynomial

$\purple{Solution\::-}$

→ x² + 1/6 x - 2

→ 6x² + x - 12 = 0

→ 6x² - 8x + 9x - 12 = 0

→ (6x² - 8x) + (9x - 12) = 0

→ 2x(3x - 4) + 3(3x + 4) = 0

→ (3x - 4) (2x + 3) = 0

→ x = 4/3 or x = -3/2

Hence, 4/3 & -3/2 are the zeroes of the polynomial