作业帮把下列各式因式分解1、（z^2-x ^2-y∧2)^ 2-4x^ 2×y∧22、x^ 3+12x-6x^2-83、x ^
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/08 04:38:32
把下列各式因式分解
1、（z^2-x ^2-y∧2)^ 2-4x^ 2×y∧2
2、x^ 3+12x-6x^2-8
3、x ^2-4xy+4y^2-x+2y-2
4、x^3+3x^2-4
1、（z^2-x ^2-y∧2)^ 2-4x^ 2×y∧2
2、x^ 3+12x-6x^2-8
3、x ^2-4xy+4y^2-x+2y-2
4、x^3+3x^2-4
1.用平方差公式:
(z²-x²-y²)²-4x²y²
=[(z²-x²-y²)-2xy][(z²-x²-y²+2xy]
=[z²-(x+y)²][z²-(x-y)²]
=(z-x-y)(z+x+y)(z-x+y)(z+x-y)
2.x³+12x-6x²-8
=x³-6x²+12x-8,可用多项式定理
=(x)³-3(x)²(2)+3(x)(2)²-(2)³
=(x-2)³
3.x²-4xy+4y²-x+2y-2
=(x-2y)²-(x-2y)-2
=u²-u-2,u=x-2y,用十字相乘法
=(u-2)(u+1)
=(x-2y-2)(x-2y+1)
4.x³+3x²-4
设f(x)=x³+3x²-4,常数-4的因子为{±1,±2,±4}
∵f(1)=1+3-4=0
∴(x-1)是其中一个因式,
用综合除法,(x³+3x²-4)/(x-1)化简得:
f(x)=(x-1)(x²+4x+4)
=(x-1)(x+2)²
∴x³+3x²-4=(x-1)(x+2)²
(z²-x²-y²)²-4x²y²
=[(z²-x²-y²)-2xy][(z²-x²-y²+2xy]
=[z²-(x+y)²][z²-(x-y)²]
=(z-x-y)(z+x+y)(z-x+y)(z+x-y)
2.x³+12x-6x²-8
=x³-6x²+12x-8,可用多项式定理
=(x)³-3(x)²(2)+3(x)(2)²-(2)³
=(x-2)³
3.x²-4xy+4y²-x+2y-2
=(x-2y)²-(x-2y)-2
=u²-u-2,u=x-2y,用十字相乘法
=(u-2)(u+1)
=(x-2y-2)(x-2y+1)
4.x³+3x²-4
设f(x)=x³+3x²-4,常数-4的因子为{±1,±2,±4}
∵f(1)=1+3-4=0
∴(x-1)是其中一个因式,
用综合除法,(x³+3x²-4)/(x-1)化简得:
f(x)=(x-1)(x²+4x+4)
=(x-1)(x+2)²
∴x³+3x²-4=(x-1)(x+2)²
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