Show $f(x)=x^ 2 - 14$ can't be factorized as $(ax + b )(c x +d)$ in $\mathbb Z_{2014}[x]$ show $f(x)=x^ 2 - 14 $ can't be expressed as $(ax + b )(c x +d)$ in $\mathbb Z_{2014}[x]$
I tried this by different  way comparing with it in my book .
Here is my attempt
I've got some idea from factorization of 2014.
Take the case, equation $f(x)= 0 $ for mod ${2014}$
If so, $f(x)\equiv 0 \pmod 2$, $f(x)\equiv 0 \pmod{ 19}$and $f(x)\equiv 0 \pmod {53}$ holds.
plus, put $f_2 \equiv f \pmod 2$,  $f_{19} \equiv f \pmod   {19} $ and vice versa $f_{53}$
Suppose the $f$ can  be factorized by those two polynomials(I.e. $f = (ax+b)(cx+d)$)
Since $ac= 1$ for mod $2014$, $gcd(a,2014) = gcd(c,2014) =1$
Therefore $ deg f_2 =deg f_{19}= deg f_{53}= 2 $ the reason why $(ax + b )(c x +d) \equiv ( [ ax ]_p + [b]_p)( [ cx]_p + [d]_p)$ for mod $p$ (Here the $p$ is a prime number)
Therefore$ f_2 , f_{19} $and $f_{53}$ are reducible over  fields $\mathbb Z_{2}$,  $\mathbb Z_{19}$ and  $\mathbb Z_{53}$  respectively. Plus their degree are $2$, so only we left examine the existence of the roots in each fields. But the $f_{19}$ doesn't have any roots in $\mathbb Z_{19}$ by Legendre symbol. (I.e. $14$ isn't quadratic residue for mod $19 $ . ) So the $  f_{19}  $ is irreducible over $\mathbb Z_{19}$. It is contradiction.
Is  my solution  is right? please verify my solution. Thanks.
You could reduce Parcly Taxel's verbiage and increase clarity as follows: Assume the factorization holds mod 2014. Then it holds mod 19. Then $(-b/a)$ is a root of $f$ mod 19, contradicting your computation of Legendre symbol QED.
You could reduce your proof's verbiage (and increase its clarity) as follows: if $x^2-14$ factors as specified over $\mathbb Z_{2014}$, since $a$ and $c$ are coprime to $2014$ there must exist exactly two values $x$ where $x^2\equiv14\bmod2014$. By the Chinese remainder theorem this is equivalent to $14$ being a quadratic residue modulo $2$, $19$ and $53$, but $14$ is a non-residue modulo $19$ and hence there is no factorisation into the given form.