Answer:
The answer is "Zeros are 0 and 4".
Step-by-step explanation:
Given equation:
[tex]\to x^5+3x^3+x^2+2x[/tex]
applying the by direct verification:
[tex]\to \phi_{0} (x^5+3x^3+x^2+2x) =0^5+3 \cdot 0^3+0^2+2 \cdot 0 \\\\[/tex]
[tex]= 0 \ mod \ 5 \\\\[/tex]
[tex]\to \phi_{1} \ (x^5+3x^3+x^2+2x) =1^5+3 \cdot 1^3+1^2+2 \cdot 1 \\\\[/tex]
[tex]= 1+3+1+2 \ mod \ 5 \\\\ = 2 \ mod \ 5[/tex]
[tex]\to \phi_{2} \ (x^5+3x^3+x^2+2x) =2^5+3 \cdot 2^3+2^2+2 \cdot 2 \\\\[/tex]
[tex]= 2+4+4+4 \ mod \ 5 \\\\ = 4 \ mod \ 5[/tex]
[tex]\to \phi_{3}\ (x^5+3x^3+x^2+2x) =3^5+3 \cdot 3^3+3^2+2 \cdot 3 \\\\[/tex]
[tex]= 3+1+4+1 \ mod \ 5 \\\\ = 4 \ mod \ 5[/tex]
[tex]\to \phi_{4} \ (x^5+3x^3+x^2+2x) =4^5+3 \cdot 4^3+4^2+2 \cdot 4 \\\\[/tex]
[tex]= 4+2+1+3 \ mod \ 5 \\\\ = 0 \ mod \ 5[/tex]
