Question

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

Answer 1

Answer:

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

Answer 2

Answer:

x=-1,1,2,3

Step-by-step explanation:

(x^2-1)-> (x-1)(x+3)-> x=1,-1

Factor x^2-5x+6-> (x-3)(x-2)-> x=2,3

Thus, x=-1,1,2,3