更新1:
okay i get thats the right answer but what method did you use? I dont understand how to do this??
Help Factoring X^5 - 1?
回答 (7)
2009-06-08 7:05 am
✔ 最佳答案
=(x-1)(x^4-x^3+x^2-x+1)
2009-06-08 6:59 am
X^5 - 1=(X - 1)(x^4+x^3+x^2+x+1)
(x^4+x^3+x^2+x+1) not factorable over real numbers.
(x^4+x^3+x^2+x+1) not factorable over real numbers.
2009-06-08 7:00 am
It's barely possible, - all you can do here is x^5(1-1/x^5).That's all, otherwise the equation is prime. At least, I don't see anything more meaningful.
2009-06-08 10:16 pm
x^5 - 1
Answer:
(x - 1)(x ^4 + x ^3 + x^2 + x + 1)
Multiply using the FOIL Method to confirm (check and test) the answer
x * x^4 = x^5
x * x^3 = x^4
x * x^2 = x^3
x * x = x^2
x * 1 = 1x
-1 * x^4 = -x^4
-1 * x^3 = -x^3
-1 * x^2 = -x^2
-1 * x = -1x
-1 * 1 = -1
x^5 + x^4 + x^3 + x^2 + 1x - x^4 - x^3 - x^2 - 1x - 1
x^5 - 1
Answer:
(x - 1)(x ^4 + x ^3 + x^2 + x + 1)
Multiply using the FOIL Method to confirm (check and test) the answer
x * x^4 = x^5
x * x^3 = x^4
x * x^2 = x^3
x * x = x^2
x * 1 = 1x
-1 * x^4 = -x^4
-1 * x^3 = -x^3
-1 * x^2 = -x^2
-1 * x = -1x
-1 * 1 = -1
x^5 + x^4 + x^3 + x^2 + 1x - x^4 - x^3 - x^2 - 1x - 1
x^5 - 1
2009-06-08 8:29 am
x^n - 1 = (x - 1)[x^(n - 1) + x^(n - 2) + x^(n - 3) + ... ... + x^2 + x + 1]
x^5 - 1
= x^5 - 1^5
= (x - 1)(x^4 + x^3 + x^2 + x + 1)
x^5 - 1
= x^5 - 1^5
= (x - 1)(x^4 + x^3 + x^2 + x + 1)
參考: Difference of two squares (2)
http://www.themathpage.com/alg/difference-two-squares-2.htm#sum
2009-06-08 7:11 am
(x^5 - 1) = (x - 1)(x^4 + x^3 + x^2 + x + 1)
2009-06-08 7:06 am
ANY (x^n - 1) = (x - 1)(x^(n-1) + x^(n-2) + x^(n-3) + ...+ 1), so
(x - 1)(x^4 + x^3 + x^2 + x + 1) (multiply and you'll see)
(x - 1)(x^4 + x^3 + x^2 + x + 1) (multiply and you'll see)
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