Answer to Question #284664 in Abstract Algebra for Bret

Question

Let f(x) = x3 + 6 be an element Z [x]. Write f(x) as a product of irreducible 7

polynomials over Z7.

Answer 1

Solution:

Consider the function stated below.

$\begin{array}{r} f(x)=x^{3}+6 \\ \in Z_{7}[x] \end{array}$

Make the observations as stated below.

$\begin{aligned} f(1) &=1+6 \\ &=7 \\ &=0(\bmod 7) \end{aligned}$

Therefore, 1 is a zero of f(x).

Similarly;

$\begin{aligned} f(2) &=2^{3}+6 \\ &=14 \\ &=7 \times 2 \\ &=0(\bmod 7) \end{aligned}$

Therefore, 2 is a zero of f(x)

Similarly;

$\begin{aligned} f(4) &=4^{3}+6 \\ &=70 \\ &=7 \times 10 \\ &=0(\bmod 7) \end{aligned}$

Therefore, 4 is a zero of f(x).

Now, since {1,2,4} is a set of zeros of f(x);

Therefore, by virtue of the Factor Theorem, conclude that;$\{(x-1),(x-2),(x-4)\}$ is a set of factors of f(x) .

This implies;

$(x-1)(x-2)(x-4) \mid x^{3}+6$

Also, note that;

$\begin{aligned} \operatorname{deg}((x-1)(x-2)(x-4)) &=3 \\ &=\operatorname{deg}\left(x^{3}+6\right) \end{aligned}$

Hence, the required solution is $x^{3}+6=(x-1)(x-2)(x-4).$