Recall we say a polynomial is monic if its leading coefficie

Recall, we say a polynomial is monic if its leading coefficient is one. Let f(x) = x^2 + 3x + 2 be a polynomial in Z_6[x]. Write f(x) as a product of two monic linear polynomials in Z_6[x]. Write f(x) as a product of two monic linear polynomials in Z_6[x], where these factors are different than those found in (a). Verify that g(x) = x^2 + 2x + 5 has no roots in Z_6. Write g(x) as a product of two non-constant polynomials in Z_6[x].

Solution

a)

x^2+3x+2=x^2+2x+x+2=(x+2)(x+1)

b)

2=-4 mod 6

So,

f(x)=x^2+3x-4=x^2+4x-x-4=(x+4)(x-1)

c)

x^2+2x+5=0 mod 6

x^2+2x+1+4=0 mod 6

(x+1)^2+4=0 mod 6

(x+1)^2=-4=2 mod 6

Now we check if 2 can be a quadratic residue modulo 6

1^2=1

2^2=4

3^2=9=3 mod 6

4^2=16=4 mod 6

5^2=1 mod 6

Hence, (x+1)^2=2 mod 6 is not possible

So,

g(x) has not roots in Z_6

It is not possible

BEcause, if g(x)=f(x)h(x) where f and h are non constant the both would have degree equal to 1

So, they would be of the form ,f(x)=x+a,h(x)=x+b

HEnce, x=-a and x=-b are roots of g(x) in Z_6

which is not possible as we just proved