Is $f$ a product of two polynomials of degree $2$ over $\Bbb Z$.

Question

Consider the polynomial $f(x)=x^4-x^3+14x^2+5x+16$.Let $\Bbb F_p$ denote the field with $p$ elements.

Which are correct?

1. Considering $f$ as a polynomial with co-efficients in $\Bbb F_3$,it has no roots in $\Bbb F_3$.

2. Considering $f$ as a polynomial with co-efficients in $\Bbb F_3$,it is a product of two irreducible polynomials of degree $2$.

3. Considering $f$ as a polynomial with co-efficients in $\Bbb F_7$,it has an irreducible factor of degree $3$ over $\Bbb F_7$.

4. $f$ is a product of two polynomials of degree $2$ over $\Bbb Z$.

In $\Bbb F_3$ $f(x)=x^4-x^3+2x^2+2x+1$; So $f(2)=21=0$

So 1 is false.

Suppose that $f(x)=(x^2+ax+b)(x^2+cx+d)\implies x^4+(a+c)x^3+(ac+b+d)x^2+x(bc+da)+db$.

We must have $bd=1\implies $ either $b=d=1$ or $b=d=2$.

On comparing co-efficients of $x$ we have $a+c=1$ and of $x^3$ we have $a+c=-1$ which is false. So $2$ is false.

For $3$,$f(x)=x^4-x^3+5x+2$ in $\Bbb F_7$ $f(x)=(x^3+5)(x-1)$ and $x^3+5$ is irreducible in $\Bbb F_7$ and so it is true.

I don't know how to solve $4$. Please help.

However correct options given are $1,2,3$. Please suggest required edits

Answer 1

In 1, you already showed that $f(2)=0$, so $f(x)$ is divisible by $(x-2)$ and cannot be the product of two irreducible quadratics.

Assume $f(x)=g(x)h(x)$. Then one of $g(x)\bmod 7$, $h(x)\bmod 7$ must be a multiple of the irreducible cubic factor of $f(x)\bmod 7$, hence one of $g(x),h(x)$ must have degree $\ge 3$ and cannot be quadratic.