求因式分解(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 07:36:48
求因式分解(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
求因式分解(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
求因式分解(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
原式=[(x^4+1)-4x²][(x^4+1)+3x²]+10x^4
=(x^4+1)²-x²(x^4+1)-12x^4+10x^4
=(x^4+1)²-x²(x^4+1)-2x^4
=(x^4+1-2x²)(x^4+1+x²)
=(x²-1)²[(x^4+2x²+1)-x²]
=(x+1)²(x-1)²[(x²+1)²-x²]
=(x+1)²(x-1)²(x²+x+1)(x²-x+1)
(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
为方便起见,可以用换元法来分解,设x^4+1=M,x²=N,
则
原式
=(M-4N)(M+3N)+10N²
=M²-MN-12N²+10N²
=M²-MN-2N²
=(M-2N)(M+N)
...
全部展开
(X^4-4X^2+1)(X^4+3X^2+1)+10X^4
为方便起见,可以用换元法来分解,设x^4+1=M,x²=N,
则
原式
=(M-4N)(M+3N)+10N²
=M²-MN-12N²+10N²
=M²-MN-2N²
=(M-2N)(M+N)
=(x^4+1-2x²)(x^4+1+x²)
=(x^2-1)²[(x^4+2x²+1)-x²]
=(x-1)²(x+1)²[(x²+1)²-x² ]
=(x-1)²(x+1)²(x²+1-x)(x²+1+x)
收起
令x^4+1=A,x^2=B
则
(x^4-4x^2+1)(x^4+3x^2+1)+10x^4
=(A-4B)(A+3B)+10B^2
=A^2-AB-12B^2+10B^2
=A^2-AB-2B^2
=(A-2B)(A+B)
=(x^4+1-2x^2)(x^4+1+x^2)
其中
x^4+1-2x^2
=(x^2-1...
全部展开
令x^4+1=A,x^2=B
则
(x^4-4x^2+1)(x^4+3x^2+1)+10x^4
=(A-4B)(A+3B)+10B^2
=A^2-AB-12B^2+10B^2
=A^2-AB-2B^2
=(A-2B)(A+B)
=(x^4+1-2x^2)(x^4+1+x^2)
其中
x^4+1-2x^2
=(x^2-1)^2
=(x-1)^2(x+1)^2
x^4+1+x^2
=x^4+2x^2+1-x^2
=(x^2+1)^2-x^2
=(x^2+1-x)(x^2+1+x)
则
(x^4-4x^2+1)(x^4+3x^2+1)+10x^4
=(x-1)^2(x+1)^2(x^2+1-x)(x^2+1+x)
收起