jmc

In order to factor the polynomial, we will try to solve the equation $x^8+98x^4+1=0.$ First, we can divide both sides by $x^4,$ to get $x^4+98+\frac{1}{x^4}=0,$ so $$x^4+\frac{1}{x^4}=-98.$$Then $$x^4+2+\frac{1}{x^4}=-96,$$which we can write as ${\left(x^2+\frac{1}{x^2}\right)}^2=-96.$ Hence, $$x^2+\frac{1}{x^2}=\pm 4i\sqrt{6}.$$Then $$x^2-2+\frac{1}{x^2}=-2\pm 4i\sqrt{6},$$which we can write as $${\left(x-\frac{1}{x}\right)}^2=-2\pm 4i\sqrt{6}.$$To work with this equation, we will find the square roots of $-2\pm 4i\sqrt{6}.$

Assume that $\sqrt{-2+4i\sqrt{6}}$ is of the form $a+b.$ Squaring, we get $$-2+4i\sqrt{6}=a^2+2ab+b^2.$$We set $a^2+b^2=-2$ and $2ab=4i\sqrt{6},$ so $ab=2i\sqrt{6}.$ Then $a^2{b}^2=-24,$ so $a^2$ and $b^2$ are the roots of the quadratic $$t^2+2t-24=0,$$which factors as $(t-4)(t+6)=0.$ Hence, $a^2$ and $b^2$ are 4 and $-6$ in some order, which means $a$ and $b$ are $\pm 2$ and $\pm i\sqrt{6}$ in some order.

We can check that $$(2+i\sqrt{6}{)}^2=4+4i\sqrt{6}-6=-2+4i\sqrt{6}.$$Similarly, $$\begin{array}{rl}(-2-i\sqrt{6}{)}^2& =-2+4i\sqrt{6},\\ (2-i\sqrt{6}{)}^2& =-2-4i\sqrt{6},\\ (-2+i\sqrt{6}{)}^2& =-2-4i\sqrt{6}.\end{array}$$Thus, $$x-\frac{1}{x}=\pm 2\pm i\sqrt{6}.$$If $$x-\frac{1}{x}=2+i\sqrt{6},$$then $$x-\frac{1}{x}-2=i\sqrt{6}.$$Squaring both sides, we get $$x^2-4x+2+\frac{4}{x}+\frac{1}{x^2}=-6,$$so $$x^2-4x+8+\frac{4}{x}+\frac{1}{x^2}=0.$$This simplifies to $x^4-4x^3+8x^2+4x+1.$

Similarly, $$x-\frac{1}{x}=-2+i\sqrt{6}$$leads to $x^4+4x^3+8x^2-4x+1.$ Thus, $$x^8+98x^4+1=(x^4+4x^3+8x^2-4x+1)(x^4-4x^3+8x^2+4x+1).$$Evaluating each factor at $x=1,$ the final answer is $(1+4+8-4+1)+(1-4+8+4+1)=\boxed{20}.$