點因式分解:
2x^3+3x^2-1 to (x+1)^2 (2 x-1)
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好 Basic o既 數學題(急!!!)
2010-02-17 10:31 pm
回答 (2)
2010-02-17 10:43 pm
✔ 最佳答案
2x^3+3x^2-1= 2x^3 + 2x^2 + x^2 - 1
= (2x^3 + 2x^2) + (x^2 - 1)
= (2x^2)(x+1) + (x-1)(x+1)
= (x+1)(2x^2 + x-1)
= (x+1)(x+1)(2x-1)
= (x+1)^2 (2 x-1)
也可用餘式定理代 - 1 入 x 計到餘式=0,得一個因式 (x+1)
再用2x^3+3x^2-1 除以 (x+1)...
2010-02-17 10:50 pm
let f(x)=2x^3+3x^2-1
f(-1)=2(-1)^3+3(-1)^2-1
=0
so that x+1is a factor of f(x)
lf(1/2)=2(1/2)^3+3(1/2)^2-1
=0
so that x-1/2is a factor of f(x)
f(x)=2x^3+3x^2-1
=(x+1)(x-1/2)(2x+2) (by long division)
=2(x+1)(x-1/2)(x+1)
=(x+1)(2x-1)(x+1)
=(x+1)^2(2x-1)
f(-1)=2(-1)^3+3(-1)^2-1
=0
so that x+1is a factor of f(x)
lf(1/2)=2(1/2)^3+3(1/2)^2-1
=0
so that x-1/2is a factor of f(x)
f(x)=2x^3+3x^2-1
=(x+1)(x-1/2)(2x+2) (by long division)
=2(x+1)(x-1/2)(x+1)
=(x+1)(2x-1)(x+1)
=(x+1)^2(2x-1)
收錄日期: 2021-04-21 22:08:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100217000051KK00598