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21 May, 07:17

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

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  1. 21 May, 09:41
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    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
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