solve for x: 2x^3+4x^2+3x+6=0?
回答 (3)
2008-06-01 3:28 pm
✔ 最佳答案
2x^3 + 4x^2 + 3x + 6 = 0(2x^2 + 3)(x + 2) = 0
2x^2 + 3 = 0
2x^2 = -3
x^2 = -3/2
x = √(-3/2) (imaginary number)
x + 2 = 0
x = -2
∴ x = -2
2008-06-01 3:35 pm
-2 is a root by trial and error. There seems to be no other roots. You may graph the function.
Divide 2x^3+4x^2+3x+6 by (x+2) and see if you can factor it.
Divide 2x^3+4x^2+3x+6 by (x+2) and see if you can factor it.
2008-06-01 3:28 pm
2x^2(x+2) + 3(x+2) = 0
(2x^2+3)(x+2) = 0
x=-2
(2x^2+3)(x+2) = 0
x=-2
收錄日期: 2021-05-01 10:34:11
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https://hk.answers.yahoo.com/question/index?qid=20080601072404AAYG736