Solving fourth degree polynomial

[j-lee00]

Here it is

(x^4)-(3x^3)-(3x^2)+2 = 0

Two solutions are

x = 1 - i
x = 1 + i

How can i find the other solutions, (not graphically)

Thanks

 

[FrogPad]

Here it is

(x^4)-(3x^3)-(3x^2)+2 = 0

Two solutions are

x = 1 - i
x = 1 + i

How can i find the other solutions, (not graphically)

Thanks

General solution:
http://en.wikipedia.org/wiki/Quartic_equation

Numerical method:
http://en.wikipedia.org/wiki/Newton's_method

 

[uart]

General solution:
http://en.wikipedia.org/wiki/Quartic_equation

Numerical method:
http://en.wikipedia.org/wiki/Newton's_method[/QUOTE]

There's no need for that. OP already has two of the roots so all that's required is to divide the original quartic by (x^2 - 2x + 2) and factorize the resultant quadratic.

BTW. (x^2 - 2x + 2) = (x-(1+i)) (x-(1-i))

 

[FrogPad]

There's no need for that. OP already has two of the roots so all that's required is to divide the original quartic by (x^2 - 2x + 2) and factorize the resultant quadratic.

BTW. (x^2 - 2x + 2) = (x-(1+i)) (x-(1-i))

good call.

 

[HallsofIvy]

One serious problem the OP has is that 1-i and 1+ i are NOT roots of
x⁴-3x³-3x²+2= 0!

If x= 1- i then x⁴-3x³-3x²+2= 4+ 12i and if x= 1+ i then x⁴-3x³-3x²+2= 4- 12i, not 0.

 

[uart]

One serious problem the OP has is that 1-i and 1+ i are NOT roots of
x⁴-3x³-3x²+2= 0!

Oh yeah you're right. I just took his word for that and didn't check it.

My guess is that OP was given the two roots as part of the problem and has made a typo error in posting the equation.

 

[j-lee00]

yes sorry it was a typo but i got the method

thanks