In $\mathbb{Z}_7[x]$ ,write $f(x)=2x^3+2x+3$ as product of irreducibles .

Question

In $\mathbb{Z}_7[x]$ ,write $f(x)=2x^3+2x+3$ as product of irreducibles . make all your coefficients either $0,1,2,3,4,5$ or $6$

My attempt: as $f(x)=2x^3+2x+3\rightarrow f(1)=2+2+3=7=0 $

so we can write $f(x)=(x-1)g(x)$

but how to find $g(x)?$

Answer 1

As I mentioned in the comments, you can just apply long division and obtain $g(x) = 2 x^2 + 2 x + 4 = 2(x^2 + x + 2)$. Again when you sub in $x = 3$, $g(3) = 0$. One method is to differentiate $g(x)$ which gives you $g(x) = 2x + 1$, so $x = 3$ is a double root. Otherwise you can just simply apply long division again and $g(x) = 2(x - 3)^2$. So you have $f(x) = 2(x - 1)(x - 3)^2$.

Answer 2

After long dividing, we have $$2x^3+2x+3=(x-1)(2x^2+2x+4)+7 $$ which shows $g(x)=2x^2+2x+4$. The rest is as a above, noting that $g(3)=0$.

Answer 3

When the degree is low we can just do some tricks to avoid long division. We have $f(1) = 7, f(3) = 63$, so these are both roots. So it splits into at least two linear polynomials, but by degree counting the other must be linear as well. So we have $$2x^3 + 2x + 3 \equiv 2(x-1)(x-3)(x-a) \pmod 7$$ for some $a$. The $x^2$ term is $2(1+3+a)$ on the right side and $0$ on the left side, so $a=3$. Thus it factorizes as $$2(x-1)(x-3)^2.$$

Answer 4

You have $2x^3+2x+3$ and you want to factor out $x-1$, so we need to find some expression in the form of $(x-1)(ax^2+bx+c)+d$ to solve it.

Let's expend this expression:$$(x-1)(ax^2+bx+c)+d=ax^3-ax^2+bx^2-bx+cx-c+d=ax^3+(b-a)x^2+(c-b)x+(d-c)$$

Now compare it to the original and you get:$$\begin{cases}a=2\\b-a=0\implies b=2\\c-b=2\implies c=4\\d-c=3\implies d=7\end{cases}$$ therefore $$2x^3+2x+3=(x-1)(2x^2+2x+4)+7=(x-1)\left[(2x^2+2x+4)+\frac7{x-1}\right]\implies g(x)=(2x^2+2x+4)+\frac7{x-1}$$