23. It is given that $a, b, c$ are three real numbers such that the roots of the equation $x^{2}+3 x-1=0$ also satisfy the equation $x^{4}+a x^{2}+b x+c=0$. Find the value of $a+b+4 c+100$.
23. Answer: 93
By long division, we have
$$
x^{4}+a x^{2}+b x+c=\left(x^{2}+3 x-1\right) \cdot\left(x^{2}-3 x+(a+10)\right)+(b-3 a-33) x+(c+a+10) \text {. }
$$

Let $m_{1}, m_{2}$ be the two roots of the equation $x^{2}+3 x-1=0$. Note that $m_{1} \neq m_{2}$, since the discriminant of the above quadratic equation is $3^{2}-4 \cdot 1 \cdot 1 \cdot(-1)=13 \neq 0$. Since $m_{1}, m_{2}$ also satisfy the equation $x^{4}+a x^{2}+b x+c=0$, it follows that $m_{1}$ and $m_{2}$ also satisfy the equation
$$
(b-3 a-33) x+(c+a+10)=0 .
$$

Thus we have
$$
(b-3 a-33) m_{1}+(c+a+10)=0,
$$
and
$$
(b-3 a-33) m_{2}+(c+a+10)=0 .
$$

Since $m_{1} \neq m_{2}$, it follows that $b-3 a-33=0$ and $c+a+10=0$. Hence we have $b=$ $3 a+33$ and $c=-a-10$. Thus $a+b+4 c+100=a+(3 a+33)+4(-a-10)+100=93$.