Question

Show that the polynomial \(x^4+x^2+1\) has no integer roots, but that it has a root modulo 3, and factorize it over \(Z_3\)

Answer 1

By the rational roots theorem the only possible roots for x^4+x^2+1 = 0 are 1 or -1, and a manual check shows neither works. If you don't want to reference the rational roots theorem, you can rewrite the equation as x^2(x^1+1) = -1, which shows x is a divisor of 1.

Clearly x = 1 and -1 are both roots mod 3, (+/-1)^4+(+/-1)^2+1 = 1+1+1 = 0 mod 3. So x^2-1 divides x^4+x^2+1 = (x^2-1)^2 mod 3.

Answer 2

i get the second half but i'm still a little confused about the first half... in the book, it only shows \(x^4+x^2+1=(x^2+2)(x^2+2)=(x+1)(x+1)(x+2)(x+2)\) which does not make any sense -_-

Answer 3

@love_jessika15 can you explain more?

Answer 4

hmm

Answer 5

You got me confused now lol

Answer 6

sorry... i'm so confused as well :(

Answer 7

@satellite73 can you help?

Answer 8

I'm sorry ):

Answer 9

it's ok and thanks

Answer 10

@jim_thompson5910 can you help me?

Answer 11

(x^2+2)(x^2+2) = x^4 + 2x^2 + 2x^2 + 4

(x^2+2)(x^2+2) = x^4 + 4x^2 + 4

when you reduce the right side mod 3, you basically take the remainders if you divide each coefficient by 3

so 1 = 1 (mod 3), 4 = 1 (mod 3) because 4/3 = 1 remainder 1

which means

(x^2+2)(x^2+2) = x^4 + 4x^2 + 4 = x^4 + x^2 + 1 (mod 3)

Answer 12

do I do the same thing to get \((x+1)^2(x+2)^2\)?