Question

2x^4-x^3-6x^2-x 2

用换元法因式分解，不要网上抄的，谢谢

Answer 1

f(x)=2x^4-x^3-6x^2-x+2
f(-1) = 2+1 -6+1 +2 =0
(2x^4-x^3-6x^2-x+2) =(x+1) (2x^3+ ax^2+bx+2)
coef. of x
2+b=-1
b=-3
coef. of x^2
b+a = -6
-3+a=-6
a=-3
(2x^4-x^3-6x^2-x+2) =(x+1) (2x^3- 3x^2-3x+2)
g(x) =2x^3- 3x^2-3x+2
g(-1) = -2-3+3+2 =0
(2x^4-x^3-6x^2-x+2) =(x+1)^2. (2x^2+cx + 2)
coef . of x
4+c=-1
c=-5
(2x^4-x^3-6x^2-x+2)
=(x+1)^2. (2x^2-5x + 2)
=(x+1)^2. (2x-1)(x-2)

追问

换元法