t³ - 2t - 15t = 0

Taking t common

t (t² - 2t - 15) = 0

By splitting the middle term

t {t² - (- 3t + 5t) - 15} = 0

t (t² + 3t - 5t - 15) = 0

t {t(t + 3) - 5(t + 3)} = 0

t (t + 3)(t - 5) = 0

⇒ t = - 3, 0, 5

Verification:

Sum of the zeroes = - (coefficient of x²) ÷ coefficient of x³

α + β + γ = - b/a

(0) + (- 3) + (5) = - (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = - 15/1

= - 15 = - 15

Product of all the zeroes = - (constant term) ÷ coefficient of x³

αβγ = - d/a

(0)(- 3)(5) = 0

0 = 0