Word Count : 547

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We are asked to find irreducible polynomials of degree 2 over ( \mathbb{F}_3 ) by looking at the factorization of ( x^9 - x ) in ( \mathbb{F}_3[x] ). Let’s solve this step by step. Over a finite field ( \mathbb{F}*q ), we have the identity: [ x^{q^n} - x = \prod*{f \text{ monic, irreducible over } \mathbb{F}_q, \deg(f) \mid n} f(x). ] Here ( q = 3 ) and ( n = 2 ), so ( x^{3^2} - x = x^9 - x ) factors into all monic irreducible polynomials over ( \mathbb{F}_3 ) of degrees dividing 2. Step 1: Factor out the linear factors (degree 1). The linear polynomials over ( \mathbb{F}_3 ) are ( x, x-1, x-2 ). They are clearly irreducible of degree 1. So the product of linear factors is: [ __________ ________ ____ ____ _____ _________.
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