8x^4 - 4x^3 - 8x^2 + 3x + 1
有無可能......?
一元四次方程.....
2010-04-12 6:48 am
回答 (3)
2010-04-12 8:21 am
x = 1
x = 0.6234898018587336
x = -0.22252093395631445
x = -0.900968867902419
x = 0.6234898018587336
x = -0.22252093395631445
x = -0.900968867902419
2010-04-12 7:07 am
By observation:
8x^4 - 4x^3 - 8x^2 + 3x + 1
= (8x^4 - 4x^3 - 4x^2) - (4x^2 - 3x - 1)
= 4x^2(2x^2 - x - 1) - (4x + 1)(x - 1)
= 4x^2(2x + 1)(x - 1) - (4x + 1)(x - 1)
= (x - 1)[4x^2(2x + 1) - (4x + 1)]
= (x - 1)(8x^3 + 4x^2 - 4x - 1)
8x^4 - 4x^3 - 8x^2 + 3x + 1
= (8x^4 - 4x^3 - 4x^2) - (4x^2 - 3x - 1)
= 4x^2(2x^2 - x - 1) - (4x + 1)(x - 1)
= 4x^2(2x + 1)(x - 1) - (4x + 1)(x - 1)
= (x - 1)[4x^2(2x + 1) - (4x + 1)]
= (x - 1)(8x^3 + 4x^2 - 4x - 1)
2010-04-12 6:52 am
8x^4 - 4x^3 - 8x^2 + 3x + 1
有無可能......?
可能什麼？？？？
有無可能......?
可能什麼？？？？
收錄日期: 2021-04-19 21:50:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100411000051KK01923