三,四次方程求解(20分)

2x^4+9x^3+12x^2+9x+10=0

Let f(x)=2x^4+9x^3+12x^2+9x+10
cause f(-2)=2*(-2)^4+9*(-2)^3+12*(-2)^2+9*(-2)+10=0

then

f(x)=(x+2)(2x^3+5x^2+2x+5)
>>f(x)=(x+2)[x^2(2x+5)+(2x+5)]
>>f(x)=(x+2)(2x+5)(x^2+1)

therefore

(x+2)(2x+5)(x^2+1)=0
>>x=-2,-5/2,x=+/-√(-1) <<if u r f3, then this one can be rejected)

so x=-2 or -5/2

40x^3+x+1=0


實數 複數

x = 0.13200054680088900000 + 0.27797919538555500000
x = 0.13200054680088900000 - 0.27797919538555500000
x ] -0.26400109360177700000

1)

2x^4 + 9x^3 + 12x^2 + 9x + 10 = 0

put x = - 2 ,

32 - 72 + 48 - 18 + 10 = 0

So , x + 2 is the factor of the equation

2x^4 + 9x^3 + 12x^2 + 9x + 10

= (x + 2)(2x^3 + 5x^2 + 2x + 5)

= (x + 2)(2x + 5)(x^2 + 1)

(x + 2)(2x + 5)(x^2 + 1) = 0

x = - 2 , - 5 / 2 , sqrt(-1)[rejected]

The solutions are - 2 and - 5 / 2

2)

40x^3 + x + 1 = 0

x = -1

x = -1

x = -1