jmc

By the Rational Root Theorem, any rational root must be $\pm 1$ or $\pm 13.$ Checking, we find that none of these values are roots, so we look for a factorization into two quadratics. Let $$x^4-4x^3+14x^2-4x+13=(x^2+Ax+B)(x^2+Cx+D).$$Expanding the right-hand side, we get $$\begin{array}{rl}& x^4-4x^3+14x^2-4x+13\\ & =x^4+(A+C)x^3+(B+D+AC)x^2+(AD+BC)x+BD.\end{array}$$Matching coefficients, we get $$\begin{array}{rl}A+C& =-4,\\ B+D+AC& =14,\\ AD+BC& =-4,\\ BD& =13.\end{array}$$We start with the equation $BD=13.$ Either $\{B,D\}=\{1,13\}$ or $\{B,D\}=\{-1,-13\}.$ Let's start with the case $\{B,D\}=\{1,13\}.$ Without loss of generality, assume that $B=1$ and $D=13.$ Then $$\begin{array}{rl}A+C& =-4,\\ 13A+C& =-4,\\ AC& =0.\end{array}$$Then $A=0$ and $C=-4,$ so the factorization is given by $$\boxed{(x^2+1)(x^2-4x+13)}.$$