Question

find all the zeros in the indicated finite field of the given polyomial with coefficients in that field x^5 3x^3 x^2 2x

Answer 1

The answer is "Zeros are 0 and 4".

Explanation:

Given equation:

`\to x^5+3x^3+x^2+2x`

applying the by direct verification:

`\to \phi_(0) (x^5+3x^3+x^2+2x) =0^5+3 \cdot 0^3+0^2+2 \cdot 0 \\\\`

`= 0 \ mod \ 5 \\\\`

`\to \phi_(1) \ (x^5+3x^3+x^2+2x) =1^5+3 \cdot 1^3+1^2+2 \cdot 1 \\\\`

`= 1+3+1+2 \ mod \ 5 \\\\ = 2 \ mod \ 5`

`\to \phi_(2) \ (x^5+3x^3+x^2+2x) =2^5+3 \cdot 2^3+2^2+2 \cdot 2 \\\\`

`= 2+4+4+4 \ mod \ 5 \\\\ = 4 \ mod \ 5`

`\to \phi_(3)\ (x^5+3x^3+x^2+2x) =3^5+3 \cdot 3^3+3^2+2 \cdot 3 \\\\`

`= 3+1+4+1 \ mod \ 5 \\\\ = 4 \ mod \ 5`

`\to \phi_(4) \ (x^5+3x^3+x^2+2x) =4^5+3 \cdot 4^3+4^2+2 \cdot 4 \\\\`

`= 4+2+1+3 \ mod \ 5 \\\\ = 0 \ mod \ 5`