Question

因式分解：4x^4-x^2+2x-1

要详细过程

Answer 1

4x^4-x^2+2x-1
=x^2(4x^2-1)+(2x-1)
=x^2(2x+1)(2x-1)+(2x-1)
=(2x-1)[x^2(2x+1)+1]
=(2x-1)(2x^3+x^2+1)
=(2x-1)(2x^3+x^2+2x-2x+1)
=(2x-1)[(2x^3-2x)+(x^2+2x+1)]
=(2x-1)[2x(x^2-1)+(x+1)^2]
=(2x-1)[2x(x+1)(x-1)+(x+1)^2]
=(2x-1){(x+1)[2x(x-1)+x+1}
=(2x-1)[(x+1)(2x^2-2x+x+1)
=（2x-1)(x+1)(2x^2-x+1)
如果楼主是在实数范围内分解，到此步可以停止；
若是复数可继续分解。

Answer 2

=x^2*(4x^2-1)+(2x-1)
=x^2*(2x-1)(2x+1)+(2x-1)
=(2x-1)(1-x^2*(2x+1)

Answer 3

4x^4-x^2+2x-1
=x^2(4x^2-1)+(2x-1)
=x^2(2x+1)(2x-1)+(2x-1)
=(2x-1)[x^2(2x+1)+1]
=(2x-1)(2x^3+x^2+1)

Answer 4

原式=（X²）²＋2X²＋1－（X²－2X＋2）＝（X²＋1）²－（X－√2）²＝（X²+1-X+√2)(X²+1+X-√2)