求解因式分解一題
X^4 - 6X^3 + 32X
我只會把X提出來 接下來不知如何做了 = =
麻煩解答者了~感謝!!
求解因式分解一題
2009-07-10 5:44 am
回答 (4)
2009-07-10 5:57 am
✔ 最佳答案
=x(X^3-6X^2+32)=X(X^3-6X^2+8X-8X+32)
=X[X(X^2-6X+8)-8(X-4)]
=X[X(X-4)(X-2)-8(x-4)]
=X[(X-4)(X^2-2X)-8(X-4)]
=X[(X-4)(X^2-2X-8)]
=X[(X-4)(X-4)(X+2)]
=X(X-4)(X-4)(X+2)
2009-07-10 6:40 am
(x^4-6x^3+32x)
=(x-4)(x^3-2x^2-8x)
=(x-4)(x+2)(x^2-4)
=(x-4)(x+2)(x+2)(x-2)
=(x+2)^2(x-2)(x-4)
我個人認為答案是這個！
=(x-4)(x^3-2x^2-8x)
=(x-4)(x+2)(x^2-4)
=(x-4)(x+2)(x+2)(x-2)
=(x+2)^2(x-2)(x-4)
我個人認為答案是這個！
2009-07-10 6:02 am
x^4-6x^3+32x
=x(x^3 - 6x^2 + 32)
=x(x^3-4x^2-2x^2+32)
=x[x^2*(x-4)-2(x^2-16)]
=x(x-4)[x^2-2(x+4)]
=x(x-4)(x^2-2x-8)
=x(x+2)(x-4)^2
=x(x^3 - 6x^2 + 32)
=x(x^3-4x^2-2x^2+32)
=x[x^2*(x-4)-2(x^2-16)]
=x(x-4)[x^2-2(x+4)]
=x(x-4)(x^2-2x-8)
=x(x+2)(x-4)^2
2009-07-10 5:48 am
x4 - 6x3 + 32x
= x(x3 - 6x2 + 32)
設f(x) = x3 - 6x2 + 32
注意f(4) = (4)3 - 6(4)2 + 32 = 0
根據餘式定理，(x - 4)是f(x)的因子
所以，x4 - 6x3 + 32x
= x(x - 4)(x2 - 2x - 8)
= x(x - 4)(x - 4)(x + 2)
= x(x + 2)(x - 4)2
= x(x3 - 6x2 + 32)
設f(x) = x3 - 6x2 + 32
注意f(4) = (4)3 - 6(4)2 + 32 = 0
根據餘式定理，(x - 4)是f(x)的因子
所以，x4 - 6x3 + 32x
= x(x - 4)(x2 - 2x - 8)
= x(x - 4)(x - 4)(x + 2)
= x(x + 2)(x - 4)2
參考: Physics king
收錄日期: 2021-04-19 14:59:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090709000010KK11664