Question

解一元四次方程X^4+4X^3+2X^2-5X=2

解一元四次方程X^4+4X^3+2X^2-5X=2

Answer 1

x^4+4x^3+2x^2-5x-2=0
x^4-x^3+5x^3-5x^2+7x^2-7x+2x-2=0
x^3(x-1)+5x^2(x-1)+7x(x-1)+2(x-1)=0
(x-1)(x^3+5x^2+7x+2)=0
(x-1)(x^3+2x^2+3x^2+6x+x+2)=0
(x-1)[x^2(x+2)+3x(x+2)+(x+2)]=0
(x-1)(x+2)(x^2+3x+1)=0
x=1,x=-2,x=(-3+√5)/2,x=(-3-√5)/2

Answer 2

先试根，可得x=1是方程的一个根
然后用多项式除法
X^4+4X^3+2X^2-5X-2除以x-1得
X^3+5X^2+7X+2
X^3+5X^2+7X+2=X^3+5X^2+6X+X+2=
X(x+2)*(X+3)+(X+2)=
(X+2)(X^2+3X+1)
所以该方程可化为(x-1)(X+2)(X^2+3X+1)=0
方程的解为
X=1或X=-2或x=(-3+√5)/2或x=(-3-√5)/2

Answer 3

(x^2+3x+1)(x+2)(x-1)=0

x=... ...