[急!]因式分解

2015-07-22 6:41 am
(x+1)(x+3)(x+4)(x+6)-280=?

回答 (4)

2015-07-22 7:11 am
✔ 最佳答案
(x + 1) (x + 3) (x + 4) (x + 6) - 280
= (x + 1)(x + 6) (x + 3)(x + 4) - 280
= (x² + 7x + 6) (x² + 7x + 12) - 280
= (x² + 7x + 9 - 3) (x² + 7x + 9 + 3) - 280
= (x² + 7x + 9)² - 3² - 280
= (x² + 7x + 9)² - 289
= (x² + 7x + 9)² - 17²
= (x² + 7x + 9 - 17) (x² + 7x + 9 + 17)
= (x² + 7x - 8) (x² + 7x + 26)
= (x - 1) (x + 8) (x² + 7x + 26)

2015-07-22 07:12:36 補充:
別解:
(x + 1) (x + 3) (x + 4) (x + 6) - 280
= (x² + 7x + 6) (x² + 7x + 12) - 280
令 x² + 7x + 6 = y , 則 x² + 7x + 12 = y + 6
= y (y + 6) - 280
= y² + 6y - 280
= (y - 14) (y + 20)
= (x² + 7x + 6 - 14) (x² + 7x + 6 + 20)
= (x² + 7x - 8) (x² + 7x + 26)
= (x - 1) (x + 8) (x² + 7x + 26)

2015-07-22 07:13:20 補充:
餘式定理解法:
設 f(x) = (x+1) (x+3) (x+4) (x+6) - 280 = (x+1) (x+3) (x+4) (x+6) - (±2)(±4) (±5) (±7) , 則
f(1) = (2) (4) (5) (7) - (±2) (±4) (±5) (±7) = 0 , 故 f(x) 有因式 x - 1。
f(-8) = (-7) (-5) (-4) (-2) - (±2) (±4) (±5) (±7) = 0 , 故 f(x) 有因式 x + 8。

2015-07-22 07:13:53 補充:
可設 (x+1) (x+3) (x+4) (x+6) - 280 = (x - 1) (x + 8) (x² + ax + b)
常數項 = 1 × 3 × 4 × 6 - 280 = - 1 × 8 × b
- 208 = - 8b , 得 b = 26。
即 (x+1) (x+3) (x+4) (x+6) - 280 = (x - 1) (x + 8) (x² + ax + 26)
令 x = 2 , 則 3 × 5 × 6 × 8 - 280 = 10 (4 + 2a + 26)
440 = 10 (2a + 30) , 得 a = 7。

2015-07-22 07:14:13 補充:
所以 (x + 1) (x + 3) (x + 4) (x + 6) - 280 = (x - 1) (x + 8) (x² + 7x + 26)。

2015-07-22 07:21:25 補充:
解題重點是注意到 1 + 6 = 3 + 4 , 於是分 (x+1)(x+6) 及 (x+3)(x+4) 兩組使之
出現同項 x² + 7x 便於分解。
2015-07-22 7:40 am
280=2*2*2*5*7=2*4*5*7
x+1=2
x+3=4
x+4=5
x+6=7

x=1 為其一解

設 (x+1)(x+3)(x+4)(x+6)-280=(x-1)(x^3+ax^2+bx+c) ... (1)
比較常數項
1*3*4*6-280= - c ---> c=208
x=2 代入 (1)
3*5*6*8-280=8+4a+2b+208 ---> 440=8+4a+2b+208 ...(2)
224=4a+2b 112=2a+b
x=3 代入 (1)
4*6*7*9-280=2*(27+9a+3b+208 ---> 616=27+9a+3b+208 ...(3)
381=9a+3b 127=3a+b
解得 a= 15, b=72
(x+1)(x+3)(x+4)(x+6)-280=(x-1)(x^3+15x^2+82x+208)
=(x-1)(x+8)(x^2+7x+26)
參考: Paul
2015-07-22 7:19 am
設 u = x² + 7x

(x + 1) (x + 3) (x + 4) (x + 6) - 280
= [(x + 1) (x + 6)] [(x + 3) (x + 4)] - 280
= (x² + 7x + 6)(x² + 7x + 12) - 280
= (u + 6)(u + 12) - 280
= u² + 18u - 208
= (u - 8)(u + 26)
= (x² + 7x - 8)( x² + 7x + 26)
= (x - 1)(x + 8)( x² + 7x + 26)
2015-07-22 7:14 am
(x² + 7x + 6)(x² + 7x + 12) - 280 之後,
平方差其實不太容易看出來,直覺做法也可以考慮令 t = x² + 7x

= (t + 6)(t + 12) - 280
= t² + 18t + 72 - 280
= t² + 18t - 208
= (t + 26)(t - 8)
= (x² + 7x + 26)(x² + 7x - 8)
= (x - 1)(x + 8)(x² + 7x + 26)


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