We want to find the zeros of this polynomial p (x) = (x^2-1) (x^2-5x+6)

see explanation

Step-by-step explanation:

To find the zeros let p (x) = 0, that is

(x² - 1) (x² - 5x + 6) = 0

Factorise each factor

x² - 1 ← is a difference of squares and factors as (x - 1) (x + 1)

x² - 5x + 6 = (x - 2) (x - 3), thus

(x - 1) (x + 1) (x - 2) (x - 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x + 1 = 0 ⇒ x = - 1

x - 2 = 0 ⇒ x = 2

x - 3 = 0 ⇒ x = 3

The zeros are x = ± 1, x = 2, x = 3