Answer :

Here, p(x) = x —x^{2} —1

To find the zeros of p(x), consider p(x) = 0

∴ x – x^{2} – 1 = 0

Taking negative common

x^{2} – x + 1 = 0

x^{2} – x + + 1 – = 0

(Adding and subtracting in order to do factorization)

+ = 0

Using the identity: (a– b) ^{2} = (a^{2} – 2ab + b^{2})

= –

The above equation cannot be true as the square of a real number cannot be negative.

⇒ No zero exists for p(x).

Hence, the number of zeros of the given polynomial is 0.

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