設f(x)=x^4+ax^2+bx+2被(x－1)^2整除，則2a+3b=? Sol f(x)=x^4+ax^2+bx+2 f’(x)=4x^3+2ax+b f(1)=1+a+b+2=0 f’(1)=4+2a+b=0 (4+2a+b)－(1+a+b+2)=0 a=－1，b=－2 or ｘ２＋２ｘ＋２ ───────────────── ｘ２－２ｘ＋１）ｘ４＋ａｘ２＋ｂｘ＋２ ）ｘ４－２ｘ３＋ｘ２ ）──────────────── ）２ｘ３＋（ａ－１）ｘ２＋ｂｘ＋２ ）２ｘ３－４ｘ２＋２ｘ ）──────────────── ）（ａ＋３）ｘ２＋（ｂ－２）ｘ＋２ ） ２ｘ２－ ４ｘ＋２ ）──────────────── ）０ So a=－1，b=－2
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