Factorize the question plz~
-2g4+2g3+g2-1
P.S.--- '4' '3' '2' after 'g' is 次方~
---plz 列出步驟
thz!!!!!!!!!!!
勁easy數學題!!!
2009-10-17 2:10 am
回答 (2)
2009-10-17 2:20 am
✔ 最佳答案
-2g4+2g3+g2-1
=-2g3(g-1)+(g2-1)
=-2g3(g-1)+(g-1)(g+1)
=(g-1)[-2g3+(g+1)]
=(g-1)(2g3-g-1)
2009-10-17 2:47 am
Let f(g) = -2g^4+2g^3+g^2-1
Possible values for testing are : 1, -1, 1/2, -1/2
f(1) = -2+2+1-1 = 0
x-1 is a factor of f(g)
f(g) = (-2g^3+g+1)(g-1)
2009-10-16 18:48:24 補充：
Let h(g) = -2g^3+g+1
Possible values for testing are : 1, -1, 1/2, -1/2
h(1) = -2+1+1 = 0
x-1 is a factor of h(g)
h(g) = (-2g^2-2g-1)(g-1) = -(2g^2+2g+1)(g-1)
-2g^4+2g^3+g^2-1 = -(2g^2+2g+1)(g-1)^2
Possible values for testing are : 1, -1, 1/2, -1/2
f(1) = -2+2+1-1 = 0
x-1 is a factor of f(g)
f(g) = (-2g^3+g+1)(g-1)
2009-10-16 18:48:24 補充：
Let h(g) = -2g^3+g+1
Possible values for testing are : 1, -1, 1/2, -1/2
h(1) = -2+1+1 = 0
x-1 is a factor of h(g)
h(g) = (-2g^2-2g-1)(g-1) = -(2g^2+2g+1)(g-1)
-2g^4+2g^3+g^2-1 = -(2g^2+2g+1)(g-1)^2
收錄日期: 2021-04-23 20:39:08
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