應該怎樣去展開??
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更新1:
另外 因式分解2x^2-x-6 又應該是怎樣?? 先謝謝
這條數應該怎樣去展開?
回答 (3)
2010-05-05 4:30 am
✔ 最佳答案
(2x+1)(2x+3)(x-1)= (4x^2 + 2x + 6x + 3)(x-1)
= (4x^2+8x+3)(x-1)
= 4x^3+8x^2+3x-4x^2-8x-3
= 4x^3+4x^2-5x-3
2x^2-x-6
= 2x^2 - 4x + 3x - 6
= 2x(x-2) + 3(x-2)
= (x-2)(2x+3)
Or by cross-method,
2x 3
x-2
_______
3x-4x = -x
So, 2x^2-x-6 = (2x+3)(x-2)
2010-05-04 20:30:58 補充：
2x 3
x -2
_________
3x-4x = -x
參考: Knowledge is power.
2010-05-09 3:30 pm
(2x+1)(2x+3)(x-1)
=(4x^2+8x+3)(x-1)
=4x^3-4x^2+8x^2-8x+3x-3
=4x^3+4x^2-5x-3
2x^2-x-6=0
x=1/2(1+(1--48)^1/2)
=1/2(1+7)
=4or-3
So,
2x^2-x-6
=(x-4)(x+3)
2010-05-09 07:32:31 補充：
2x^2-x-6=0
1.........2
2.........-3
So,
2x^2-x-6
=(x-2)(2x+3)
=(4x^2+8x+3)(x-1)
=4x^3-4x^2+8x^2-8x+3x-3
=4x^3+4x^2-5x-3
2x^2-x-6=0
x=1/2(1+(1--48)^1/2)
=1/2(1+7)
=4or-3
So,
2x^2-x-6
=(x-4)(x+3)
2010-05-09 07:32:31 補充：
2x^2-x-6=0
1.........2
2.........-3
So,
2x^2-x-6
=(x-2)(2x+3)
參考: me, me
2010-05-05 6:47 am
(2x+1)(2x+3)(x-1)
(2x(2x+3)+1(2x+3))(x-1)
(4x^2+6x+2x+3)(x-1)
(4x^2+8x+3)(x-1)
x(4x^2+8x+3)-1(4x^2+8x+3)
4x^3+8x^2+3x-4x^2-8x-3
4x^3+4x^2-5x-3
2x^2-x-6
(2x+3)(x-2)
(2x(2x+3)+1(2x+3))(x-1)
(4x^2+6x+2x+3)(x-1)
(4x^2+8x+3)(x-1)
x(4x^2+8x+3)-1(4x^2+8x+3)
4x^3+8x^2+3x-4x^2-8x-3
4x^3+4x^2-5x-3
2x^2-x-6
(2x+3)(x-2)
收錄日期: 2021-04-23 23:21:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100504000051KK01368