# Thread:

1. ## Polynominal root and solution problem

Hi i am trying to find the roots of the following equation,

2x4 -8x3 +13x2 -13x + 6 = 0

I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.

Regards Maria

2. Originally Posted by maria88
Hi i am trying to find the roots of the following equation,

2x4 -8x3 +13x2 -13x + 6 = 0

I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.

Regards Maria
Since you have found the solutions x=1 and x=2 you can factor out (x-1)(x-2) on the left side of your equation (do a polynomial division of the left side by $(x-1)(x-2)$, i.e. be $x^2-3x+2$). The remaining solutions are zeros of the remaining quadratic factor $2x^2-2x+3$ (and you know a formula to determine those).

3. Originally Posted by maria88
Hi i am trying to find the roots of the following equation,

2x4 -8x3 +13x2 -13x + 6 = 0

I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.

Regards Maria
Consider the function p(x) = 2x^4 - 8x^3 + 13x^2 - 13x + 6.

(x - 1)(x - 2) = x^2 - 3x + 2 is a factor of p(x). Divide it into p(x) to get the other quadratic factor. And you should be able to get the complex factors of this quadratic factor.

4. I would do it slightly differently. Rather than multiplying (x- 1)(x- 2) I would use "Synthetic Division" to divide by x- 1 and than again use synthetic division to divide the result by x- 2. That leaves a quadratic equation which can be solved by the quadratic formula.

5. Let the four roots be 1, 2, a+ib and a-ib.
Sum of the roots = 8/2 = 4 = 3 + 2a.
So a = 1/2.
The product of roots = 2(a^2 + b^2) = 6/2 = 3
a^2 + b^2 = 3/2
b^2 = 3/2-1/4 = 5/4
So b = sqrt(5)/2
So the roots are 1, 2, 1/2 + i*sqrt(5)/2 and 1/2 - i*sqrt(5)/2
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