Question

因式分解（x²+x）²-8x²-8x+12和 x³-2x+1求数学高手解答，

如题。。。。30分别嫌少了可以吗。。。

Answer 1

（x²+x）²-8x²-8x+12=(x+2)*(x-1)*(x+3)*(x-2)此题用换元法，设（x²+x）=t,原式=t^2-8t+12
x³-2x+1=（x-1）(x^2+x-1)此题用配凑发x³-2x+1=x^3+x^2-x-x^2-x+1=(x^3+x^2+x)-(x^2-x-1)
=x*(x^2-x-1)-(x^2-x-1)
=(x-1）(x^2+x-1)

Answer 2

(x²+x）²-8x²-8x+12=(x^2+x)^2-8(x^2+x)+12
=(x^2+x-2)(x^2+x-6)
=(x+2)(x-1)(x+3)(x-2)

x³-2x+1=x^3-x^2+x^2-x-x+1
=(x-1)(x^2+x-1)

Answer 3

解：
（x²+x）²-8x²-8x+12
=（x²+x）²-8（x²+x）²+12
=（x²+x-6）（x²+x-2）
=（x+3）(x-2)(x+2)(x-1)

x³-2x+1
=x³-1-2(x-1)
=(x-1)(x²+x+1)-2(x-1)
=(x-1)(x²+x-1)

Answer 4

（x²+x）²-8x²-8x+12
=(x²+x)-8(x²+x)+12
=(x²+x-2)(x²+x-6)
=(x-1)(x+2)(x-2)(x+3)

x³-2x+1
=(x³-x)-(x-1)
=x(x²-1)-(x-1)
=x(x-1)(x+1)-(x-1)
=(x-1)(x²+x-1)

Answer 5

（x²+x）²-8x²-8x+12
=（x²+x）²-8(x²+x）+12
=【（x²+x）-2】【（x²+x）-6】
=（x+2）（x-1）（x+3）（x-2）

x³-2x+1
=（x³-x²）+（x²-2x+1）
=x²（x-1）+（x-1）²
=（x-1）（x²+x-1）