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

Answer:

The answer is "Zeros are 0 and 4".

Step-by-step explanation:

Given equation:

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

applying the by direct verification:  

$\to \phi_{0} (x^5+3x^3+x^2+2x) =0^5+3 \cdot 0^3+0^2+2 \cdot 0$

                                         $= 0 \ mod \ 5$

$\to \phi_{1} \ (x^5+3x^3+x^2+2x) =1^5+3 \cdot 1^3+1^2+2 \cdot 1$

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

$\to \phi_{2} \ (x^5+3x^3+x^2+2x) =2^5+3 \cdot 2^3+2^2+2 \cdot 2$

                                          $= 2+4+4+4 \ mod \ 5 \\\\ = 4 \ mod \ 5$

$\to \phi_{3}\ (x^5+3x^3+x^2+2x) =3^5+3 \cdot 3^3+3^2+2 \cdot 3$

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

$\to \phi_{4} \ (x^5+3x^3+x^2+2x) =4^5+3 \cdot 4^3+4^2+2 \cdot 4$

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