9. Given the polynomial with integer coefficients
$$
f(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+a_{3} x^{2}+a_{4} x+a_{5}
$$

If $f(\sqrt{3}+\sqrt{2})=0, f(1)+f(3)=0$, then $f(-1)=$ $\qquad$
Rongshi 24.
Analysis Let $x_{0}=\sqrt{3}+\sqrt{2}$, then $x_{0}-\sqrt{3}=\sqrt{2},\left(x_{0}-\sqrt{3}\right)^{2}=2$, so $x_{0}^{2}-2 \sqrt{3} x_{0}+3=2,2 \sqrt{3} x_{0}=x_{0}^{2}+1$, thus $\left(2 \sqrt{3} x_{0}\right)^{2}=\left(x_{0}^{2}+1\right)^{2}, x_{0}^{4}-10 x_{0}^{2}+1=0$, so $x_{0}=\sqrt{3}+\sqrt{2}$ is a root of the polynomial $g(x)=x^{4}-10 x^{2}+1$. Also, $x_{0}=\sqrt{3}+\sqrt{2}$ cannot be a zero of a cubic or quadratic polynomial with integer coefficients, so $g(x)$ divides $f(x)$, so
$$
f(x)=g(x)(x-r)=\left(x^{4}-10 x^{2}+1\right)(x-r)
$$
$r$ is an integer, so $f(1)=-8(1-r)=-8+8 r, f(3)=-8(3-r)=-24+8 r$.
By $f(1)+f(3)=0$, we get $(-8+8 r)+(-24+8 r)=0, r=2$, so
$$
f(x)=\left(x^{4}-10 x^{2}+1\right)(x-2), f(-1)=24
$$