第四題,
假設題目的 f(x),\; g(x) 如下方 bugmens 所述,則
題目給的四次式 g(x)=x^4+3x^3-x^2-5x+2 被三次式 f(x)=x^3+2x^2-3x-1 除,所得之商式為 Q(x),餘式為 -x+3,
則 g(x)=f(x)Q(x)+(-x+3) 且
\displaystyle g(\alpha)=f(\alpha)Q(\alpha)+(-\alpha+3)=3-\alpha,\;g(\beta)=f(\beta)Q(\beta)+(-\beta+3)=3-\beta,\;g(\gamma)=f(\gamma)Q(\gamma)+(-\gamma+3)=3-\gamma.
因此,
\displaystyle g(\alpha)\,g(\beta)\,g(\gamma)=(3-\alpha)(3-\beta)(3-\gamma)=f(3)=35,
且
\displaystyle \frac{1}{g(\alpha)}+\frac{1}{g(\beta)}+\frac{1}{g(\gamma)}=\frac{1}{3-\alpha}+\frac{1}{3-\beta}+\frac{1}{3-\gamma}
\displaystyle \qquad \qquad \qquad =\frac{(3-\alpha)(3-\beta)+(3-\beta)(3-\gamma)+(3-\gamma)(3-\alpha)}{(3-\alpha)(3-\beta)(3-\gamma)}
\displaystyle \qquad \qquad \qquad =\frac{27-6(\alpha+\beta+\gamma)+(\alpha\beta+\beta\gamma+\gamma\alpha)}{(3-\alpha)(3-\beta)(3-\gamma)}
\displaystyle \qquad \qquad \qquad =\frac{27-6(-2)+(-3)}{35}
\displaystyle \qquad \qquad \qquad =\frac{36}{35}.