Given that x – √5 is a factor of the polynomial x³ – 3√5x² – 5x + 15√5. Find all the zeroes of the polynomial.

Let p(x) = x³ – 3√5x² – 5x + 15√5

As p(x) is cubic polynomial it will have 3 zeroes

As (x – √5) is a factor of p(x), p(√5) will be 0 by remainder theorem hence one zero of p(x) is √5

Divide p(x) by (x – √5)

Using dividend = divisor × quotient + remainder

⇒ p(x) = (x – √5)(x² – 2√5x – 15) + 0

Remaining two zeroes of p(x) we can get by equating x² – 2√5x – 15 = 0

⇒ x² – 2√5x – 15 = 0

⇒ x² – 3√5x + √5x – 15 = 0

⇒ x(x – 3√5) + √5(x – 3√5) = 0

⇒ (x + √5)(x – 3√5) = 0

⇒ x + √5 = 0 and x – 3√5 = 0

⇒ x = –√5 and x = 3√5

Hence all zeroes of x³ – 3√5x² – 5x + 15√5 are x = √5, x = –√5 and x = 3√5