Solving fourth degree polynomial

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The fourth degree polynomial equation (x^4)-(3x^3)-(3x^2)+2 = 0 initially presented two supposed solutions, x = 1 - i and x = 1 + i. However, it was pointed out that these values do not satisfy the equation, indicating a possible typo in the original problem. To find the correct solutions, the polynomial can be divided by (x^2 - 2x + 2), which corresponds to the incorrect roots, and the resulting quadratic can be factored. The discussion emphasizes the importance of verifying roots before proceeding with polynomial division. Ultimately, the correct approach to solving the polynomial requires clarification of the initial equation.
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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
 
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j-lee00 said:
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
 
FrogPad said:
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))
 
Last edited by a moderator:
uart said:
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.
 
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.
 
HallsofIvy said:
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.
 
yes sorry it was a typo but i got the method

thanks
 

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