-2q^4+2q^3+q^2-1
可否step by step 教我呢 ?
factorize數 很苦
2011-10-15 1:26 am
回答 (4)
2011-10-15 2:05 am
✔ 最佳答案
－2q^4+2q^3+q^2－1=(－2q^4+2q^3)+(q^2－1)
=－2q^3(q－1)+(q+1)(q－1)
=(q－1)(－2q^3+q+1)
=－(q－1)(2q^3－q－1)
=－(q－1)[(2q^3－2q^2)+(2q^2－2q)+(q+1)]
=－(q－1)[2q^2(q－1)+2q(q－1)+(q+1)]
=－(q－1)(2q^2+2q+1)(q－1)
=－(q－1)^2(2q^2+2q+1)
2011-10-19 1:51 am
Question : -2q^4+2q^3+q^2-1
= -2q^3(q-1)+(q-1)(q+1)
= (-2q^3+q-1)(q-1)
===========
= -2q^3(q-1)+(q-1)(q+1)
= (-2q^3+q-1)(q-1)
===========
2011-10-16 2:22 am
-2q^4+2q^3+q^2-1
=-2q^3(q-1)+(q-1)(q+1)
=(q-1)[-2q^3+(q+1)]
=(q-1)(-2q^3+q+1)
=-(q-1)(2q^3-q-1)
=(1-q)(2q^3-q-q+q-1)
=(1-q)(2q^3-2q+q-1)
=(1-q)[2q(q-1)+(q-1)]
=(1-q)[(q-1)(2q+1)
=-(q-1)^2(2q+1)
=-2q^3(q-1)+(q-1)(q+1)
=(q-1)[-2q^3+(q+1)]
=(q-1)(-2q^3+q+1)
=-(q-1)(2q^3-q-1)
=(1-q)(2q^3-q-q+q-1)
=(1-q)(2q^3-2q+q-1)
=(1-q)[2q(q-1)+(q-1)]
=(1-q)[(q-1)(2q+1)
=-(q-1)^2(2q+1)
2011-10-15 2:03 am
pls note that
q-2q^3+1 = (q-1)(-2q^2 - 2q - 1)
q-2q^3+1 = (q-1)(-2q^2 - 2q - 1)
收錄日期: 2021-05-02 10:46:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111014000051KK00392