在复数范围内分解因式 (1) x^4-4y^4 (2) (-1/2)x^2+x-3

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 23:43:57
在复数范围内分解因式 (1) x^4-4y^4 (2) (-1/2)x^2+x-3
xRJ00m#ba/Rvڛ".{P\PУAFɲaiҴE{2o2ӤݳzZݼ/UM/l~zP@rVDU9 @E 8cȒBnê HZUD6#Vw5Ê,4U5uRtɩo )o]]Br|!%$0 '8k Nn~09)pal}N,2%0Gf&~n6 6Zul="v%iF]."QR&vcV׭ͪ) {;7H

在复数范围内分解因式 (1) x^4-4y^4 (2) (-1/2)x^2+x-3
在复数范围内分解因式 (1) x^4-4y^4 (2) (-1/2)x^2+x-3

在复数范围内分解因式 (1) x^4-4y^4 (2) (-1/2)x^2+x-3
1.
x^4-4y^4
=(x^2+2y^2)(x^2-2y^2)
=(x+根号2y i)(x-根号2 yi)(x+根号2 y)(x-根号2 y)
2.
-1/2x^2+x-3
=-1/2(x^2-2x+6)
=-1/2[(x-1)^2+5]
=-1/2(x-1+根号5 i)(x-1-根号5 i)

1、原式=(x²)²-(2y²)²
=(x²+2y²)(x²-2y²)
=(x+2^½yi)(x-2^½yi)(x+2^½y)(x-2^½y)
2、原式=-½(x²-2x+6)
=-½(x+1+5^½i)(x+1-5^½i)

x^4-4y^4
=(x^2+2y^2)(x^2-2y^2)
=(x+√2yi)(x-√2yi)(x+√2y)(x-√2y)
(-1/2)x^2+x-3
=(-1/2)(x^2-2x+6)
=(-1/2)[(x-1)^2+5]
=(-1/2)(x-1+√5i)(x-1-√5i)