Question #284664

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

polynomials over Z7.


1
Expert's answer
2022-01-04T18:10:56-0500

Solution:

Consider the function stated below.

f(x)=x3+6Z7[x]\begin{array}{r} f(x)=x^{3}+6 \\ \in Z_{7}[x] \end{array}

Make the observations as stated below.

f(1)=1+6=7=0(mod7)\begin{aligned} f(1) &=1+6 \\ &=7 \\ &=0(\bmod 7) \end{aligned}

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

Similarly;

f(2)=23+6=14=7×2=0(mod7)\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;

f(4)=43+6=70=7×10=0(mod7)\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;{(x1),(x2),(x4)}\{(x-1),(x-2),(x-4)\} is a set of factors of f(x) .

This implies;

(x1)(x2)(x4)x3+6(x-1)(x-2)(x-4) \mid x^{3}+6

Also, note that;

deg((x1)(x2)(x4))=3=deg(x3+6)\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 x3+6=(x1)(x2)(x4).x^{3}+6=(x-1)(x-2)(x-4).


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