min 7b+89 = -3
b = 5
9. 已知f(x) = x' +ax²+bx+c的圖形過三點(1,0),(0,0),(-1,0),則(x-2) f(x-2)<0的解為
$5803-17
f(x) = x² + px
0 = -1-p p=-|
(7-2)-(x-2) ³-(x-2) < o
(x-2)(x-3) (x+x-1)
-1± √√5
-1=√1+4
2
(x-2)(x-3Xx-(1+√5)][*-(-+-√5) <0
othes +++ 3x11 即f(r)除以了−2x+2的餘式為
2
-7x+x0
