***数学高手请进***《2x^4-16x^3+39x^2-31x+12=0》

2x^2(x^2-8x+16)+7x^2-31x+12=0
2x^2(x-4)^2+7(x^2-31x/7+31/14)-7*(31/14)^2+12=0
2x^2(x-4)^2+7(x-31/14)^2-7*(31/14)^2+12=0
2x^2(x-4)^2+7(x-31/14)^2-7*(31/14)^2+12=0
2x^2(x-4)^2+7(x-31/14)^2+12-7*(31/14)^2=0
前两项大于0,常数项小于0,
所以有解.

1.
2x^4-16x^3+39x^2-31x+12=0
x^4-8x^3+39x^2/2-31x/2+6=0

x^2(x^2-4x+4)-4x^3-4x^2+39x^2/2-31x/2+6=0
x^2(x-2)^2-4x^3+31x^2/2-31x/2+6=0
x^2(x-2)^2-4x(x^2-4x+4)-16x^2+16x+31x^2/2-31x/2+6=0
x^2(x-2)^2-4x(x-2)^2-x^2/2+x/2+6=0
(x^2-4x)(x-2)^2-(x^2-x-12)/2=0
x(x-4)(x-2)^2-(x+3)(x-4)/2=0
(x-4)[x(x-2)^2-(x+3)/2]=0
(x-4)(x^3-4x^2+4x-x/2-3/2)=0
(x-4)(x^3-4x^2+7x/2-3/2)=0
(x-4)[x(x^2-6x+9)+6x^2-9x-4x^2+7x/2-3/2)=0
(x-4)[x(x-3)^2+(4x^2-11x-3)/2]=0
(x-4)[x(x-3)^2+(4x+1)(x-3)/2]=0
(x-4)(x-3)[x(x-3)+(4x+1)/2]=0
(x-4)(x-3)(x^2-x+1/2)=0
(x-4)(x-3)[(x-1/2)^2+1/4]=0
[(x-1/2)^2+1/4]恒大于0
所以(x-4)(x-3)=0
x=3或4

2.
2x^4-16x^3+3