1.因式分解12x^3+10x^2+2
ans=2(x+1)(6x^2-x+1)
2.設P(x)=2x^3+(2a+3)x^2-x+2。當P(x)除以x+a時,所得的餘數是當P(x)除以2x-1時所得的餘數之兩倍。求a的值
ans=±1
3.設f(x)=ax^3-x^2+bx-5及g(x)=9x^3-bx^2-ax-1。當f(x)除以3x-2時,所得餘數是-1 ; 當g(x)除以3x+1時,所得的餘數是2/9。
(a)求a和b的值 (ans= a=6,b=4)
(b)求當f(x)-g(x)除以x+2時所得的餘數 (ans=12)
我想知點計..thx
多項式因式分解
2011-06-26 5:23 pm
回答 (4)
2011-06-26 6:51 pm
✔ 最佳答案
1. 12x^3 + 10x^2 + 2 = 2(6x^3 + 5x^2 + 1)設 f(x) = 6x^3 + 5x^2 + 1, 則 f(-1) = -6 + 5 + 1 = 0
根據因式定理, x + 1是f(x)的因式。
利用長除法, 得 f(x) = (x + 1)(6x^2 - x + 1)
所以 12x^3 + 10x^2 + 2 = 2(x + 1)(6x^2 - x + 1)
2. P(1/2) = 2(1/2)^3 + (2a + 3)(1/2)^2 - 1/2 + 2 = 5/2 + a/2
P(-a) = -2a^3 + (2a + 3)a^2 + a + 2 = 3a^2 + a + 2
根據題意, P(-a) = 2P(1/2)
3a^2 + a + 2 = 5 + a
a^2 = 1
a = 1 或 -1
3. f(2/3) = -1
a(2/3)^3 - (2/3)^2 + b(2/3) - 5 = -1
4a + 9b = 60--------------------(1)
g(-1/3) = 2/9
9(-1/3)^3 -b(-1/3)^2 - a(-1/3) - 1 = 2/9
3a - b = 14-----------------------(2)
解 (1) 及 (2), 得 a = 6, b = 4
設 h(x) = f(x) - g(x)
當 h(x) 除以x+2時的餘數
= h(-2)
= f(-2) - g(-2)
= 6(-2)^3 - (-2)^2 + 4(-2) - 5 - [9(-2)^3 - 4(-2)^2 - 6(-2) - 1]
= 12 , 亦即f(x) - g(x) 除以 x + 2 時的餘數。
參考: me
2011-06-27 10:59 pm
thx everyone
2011-06-26 7:20 pm
1. 12x^3+10x^2+2
= 2(6x^3+5x^2+1)
= 2(6x^3 + 6x^2 - x^2 + 1)
= 2(6x^2(x+1) -(x^2 - 1))
= 2(6x^2(x+1) -(x-1)(x+1))
= 2((x+1)(6x^2 -(x-1))
= 2(x+1)(6x^2 - x + 1)
= 2(6x^3+5x^2+1)
= 2(6x^3 + 6x^2 - x^2 + 1)
= 2(6x^2(x+1) -(x^2 - 1))
= 2(6x^2(x+1) -(x-1)(x+1))
= 2((x+1)(6x^2 -(x-1))
= 2(x+1)(6x^2 - x + 1)
參考: use the answer as the guideline
2011-06-26 5:32 pm
Sol
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12x^3+10x^2+2
=(x+1)(12x^2－2x+2)
=2(x+1)(6x^2－x+1)
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12x^3+10x^2+2
=(x+1)(12x^2－2x+2)
=2(x+1)(6x^2－x+1)
收錄日期: 2021-04-13 18:04:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110626000051KK00188