Question

因式分解问题

x^4+5x^3+5x^2-5x+6

Answer 1

如果题目是【x^4+5x^3+5x^2-5x-6】的话：

x^4+5x^3+5x^2-5x-6
= (x^4-1)+(5x^3-5)+(5x^2-5)-(5x-5)
= (x^2+1)(x+1)(x-1)+5(x-1)(x^2+x+1)+5(x+1)(x-1)-5(x-1)
= (x-1) { (x^2+1)(x+1) + 5(x^2+x+1) + 5(x+1) - 5 }
= (x-1) { x^3+x^2+x+1 + 5x^2+5x+5 + 5x+5 - 5 }
= (x-1) { x^3+6x^2+11x+6 }
= (x-1) { (x^3+6x^2+9x) + (2x+6) }
= (x-1) { x (x+3)^2 + 2(x+3) }
= (x-1) (x+3) { x(x+3) +2 }
= (x-1) (x+3) { x^2+3x +2 }
= (x-1) (x+3) (x+2)(x+1)

追问

问一句，看出来的还是有方法？

追答

做的多了，凭感觉。
像x^4+5x^3+5x^2-5x-6，这种情况，四次项、三次项、二次项、一次项、常数项的系数分别是1，+5，+5，-5，-(5+1)，首先联想到的就是因式x-1

Answer 2

列举几个因式分解中利用到简便方法的例子以及因式分解中巧妙解题的例子~~ 明天搞定~ 1. (a+b)(a-b)^2-(a+b)^3 =(a+b)[(a-b)^2-(a+

Answer 3

x^4+5x^3+5x^2-5x+6
= (x^4-1)+(5x^3-5)+(5x^2-5)-(5x-5)+12
= (x^2+1)(x+1)(x-1)+5(x-1)(x^2+x+1)+5(x+1)(x-1)-5(x-1)+12
= (x-1) { (x^2+1)(x+1) + 5(x^2+x+1) + 5(x+1) - 5 }+12
= (x-1) { x^3+x^2+x+1 + 5x^2+5x+5 + 5x+5 - 5 }+12
= (x-1) { x^3+6x^2+11x+6 }+12
= (x-1) { (x^3+6x^2+9x) + (2x+6) }+12
= (x-1) { x (x+3)^2 + 2(x+3) }+12
= (x-1) (x+3) { x(x+3) +2 }+12
= (x-1) (x+3) { x^2+3x +2 }+12
= (x-1) (x+3) (x+2)(x+1) +12