Zeros of a polynomial

1. May 17, 2008

ehrenfest

[SOLVED] zeros of a polynomial

1. The problem statement, all variables and given/known data
Find the zeros of the polynomial

$$P(x) = x^4-6x^3+18x^2-30x+25$$

knowing that the sum of two of them is 4.

2. Relevant equations
http://en.wikipedia.org/wiki/Viète's_formulas

3. The attempt at a solution
Let x_1,x_2,x_3,x_4 be the complex roots and let x_1 +x_2 = 4. Here are the Viete relations in this case:

$$x_1+x_2+x_3+x_4 = 6$$

$$x_1 x_2 +x_1 x_3 +x_1 x_4 + x_2 x_3 + x_2 x_4 +x_3 x_4 = 18$$

$$x_1 x_2 x_3 + x_1 x_3 x_4 +x_2 x_3 x_4 +x_1 x_2 x_4= 30$$

$$x_1 x_2 x_3 x_4 = 25$$

The first one implies that x_3 +x_4 =2. And then the second one implies that x_1 x_2 + x_3 x_4 = 10 but that is as far as I can get.

2. May 17, 2008

tiny-tim

Hint: so x_2 = 4 - x_1 and x_4 = 2 - x_3.

Go substitute!

3. May 17, 2008

ehrenfest

Do you mean into

$$P(x) = x^4-6x^3+18x^2-30x+25$$

?

4. May 17, 2008

tiny-tim

No … I mean substitute into your Viete relations.

5. May 17, 2008

ehrenfest

Sorry tiny-tim, I don't know see why that helps:

For the second Viete relation I get:

$$4x_1 - 2 x_3 - x_1^2-x_3 ^2+ x_1 x_3 = 10$$

For the third one I get

$$8x_1 + 8x_3 - 2 x_1 ^2 - 4 x_3^2 = 30$$

For the fourth one I get

$$8 x_1 x_3 + x_1 ^2 x_ 3^2 - 2x_1 ^2 x_3 - 4 x_1 x_3 ^2 = 25$$

Is there a simple way to solve these equations?

6. May 17, 2008

tiny-tim

hmm … turned out more complicated than I thought.

Well … that's what happens in exams sometimes … you try something, and it doesn't work, so you try the next most obvious thing …

Now this will work:

in your
put (x_1 + x_2)s together, and (x_3 + x_4)s, and you should get two equations in x_1x_2 and x_3x_4, from which you get x_1x_2 = … and x_3x_4 = … ?