6. Discuss the factorization of the following polynomials over the integers using the prime factors of integers:
(1) $f(x)=x^{4}-3 x^{2}+9$;
(2) $f(x)=x^{4}-6 x^{3}+7 x^{2}+6 x-80$.
6. (1) Let $x=10$, then $f(10)=10^{4}-3 \cdot 10^{2}+9=9709=7 \cdot 19 \cdot 73$, since $7=10-3=x-3, 19=$ $10+9=20-1$, we have $19=x+9$ or $2x-1$. However, neither $x-3$ nor $x+9$ can divide the original expression, and in $2x-1$, the coefficient of $x$ is 2, while the highest degree term in the original expression does not contain a factor of 2, so neither of these expressions is a factor of $f(x)$. Noting that $7 \cdot 19=133$, we see that $x^{2}+3x+3$ can divide $f(x)$, hence $f(x)=\left(x^{2}+3x+3\right)\left(x^{2}-3x+3\right)$.
(2) Let $x=10$, then $f(10)=10^{4}-6 \cdot 10^{3}+7 \cdot 10^{2}+6 \cdot 10-80=4680=2^{8} \cdot 5 \cdot 3^{2} \cdot 13$, since among the factors $x-8, x-5, x-7, x+3$ or $2x-7$, only $x-5$ can divide $f(x)$, then $f(x)=(x-5)\left(x^{3}-x^{2}+2x+16\right)$. Also, $x+2$ can divide $x^{3}-x^{2}+2x+16$, so $f(x)=(x-5)(x+2)\left(x^{2}-3x+8\right)$.