Page 13 - ePD310_技術高中數學C第二冊學習講義含解析本_課本PDF

Chapter 1 式的運算 7

老師導引 範例 2 解析 P.2 學生演練

3
2
2
ணՇεධό f (x) = x + 4x - 2 ண f (x) = (a + 2)x - 4x + (2c - 1)d
2
2
3
ʿ g(x) = ax + bx + cx + ddʊ g(x) = 6x + (b + 2)x - 3d߰ f (1) = g(1)e
ٝ f (11) = g(11)df (13) = g(13)d f (-2) = g(-2)df (0) = g(0)d༊Ӌ aebe
f (17) = g(17)df (19) = g(19)dӋ aebe c ٙ࠽މ f
ced ٙ࠽މ f 答

Tips
f (x) = g(x) + f (x) - g(x) = 0 + 每一對應項的係數
相等。

答

老師導引 範例 3 解析 P.2 學生演練

2
ண aebec މྼᅰd߰ f (x) = ax + bx + c ண aebeced މྼᅰd߰
2
3
တԑ f (1) = f (2) = f (3) = 4dۆ aebec f (x) = (a - 2)x + (b + 1)x + cx + (3d - 2)
ٙ࠽މ f တԑ f (1) = f (2) = f (3) = f (5) = 4dۆ ae
Tips beced ٙ࠽މ f
f (x) = g(x) + f (x) - g(x) = 0 + 每一對應項的係數 答
相等。

答

重點 三 多項式的四則運算

1. Շεධό޴̋dఱ݊ҪΝϣධٙڷᅰ޴̋iϾ޴ಯࣛdఱ݊ҪΝϣධٙڷᅰ޴ಯf

n
註 ax ! bx = (a ! b)x n
n
2. ຅ f (x) + g(x) ! 0 ࣛddeg (f (x) + g(x)) ʃ׵אഃ׵ deg f (x) ၾ deg g(x) ʕٙ༰ɽᅰi
຅ f (x) - g(x) ! 0 ࣛddeg (f (x) - g(x)) ʃ׵אഃ׵ deg f (x) ၾ deg g(x) ʕٙ༰ɽᅰf