Finding the Zeros of a Polynomial: Viète's Formulas [SOLVED]

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In summary, the problem is to 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. Using Viete's formulas, it can be shown that x_1 + x_2 = 4 and x_3 + x_4 = 2. Substituting these values into the Viete relations, it can be simplified to two equations in x_1x_2 and x_3x_4. Solving these equations will give the values of x_1x_2 and x_3x_4, which can then be used to find the zeros of the polynomial
  • #1
ehrenfest
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[SOLVED] zeros of a polynomial

Homework Statement


Find the zeros of the polynomial

[tex]P(x) = x^4-6x^3+18x^2-30x+25[/tex]

knowing that the sum of two of them is 4.

Homework Equations


http://en.wikipedia.org/wiki/Viète's_formulas

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:

[tex]x_1+x_2+x_3+x_4 = 6 [/tex]

[tex] 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 [/tex]

[tex] 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 [/tex]

[tex] x_1 x_2 x_3 x_4 = 25[/tex]

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.

Please just give a hint.
 
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  • #2
ehrenfest said:
x_1 +x_2 = 4 … x_3 +x_4 =2.

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

Go substitute! :smile:
 
  • #3
Do you mean into

[tex]
P(x) = x^4-6x^3+18x^2-30x+25
[/tex]

?
 
  • #4
No … I mean substitute into your Viete relations. :smile:
 
  • #5
Sorry tiny-tim, I don't know see why that helps:

For the second Viete relation I get:

[tex]4x_1 - 2 x_3 - x_1^2-x_3 ^2+ x_1 x_3 = 10[/tex]

For the third one I get

[tex]8x_1 + 8x_3 - 2 x_1 ^2 - 4 x_3^2 = 30[/tex]

For the fourth one I get

[tex]8 x_1 x_3 + x_1 ^2 x_ 3^2 - 2x_1 ^2 x_3 - 4 x_1 x_3 ^2 = 25[/tex]

Is there a simple way to solve these equations?
 
  • #6
hmm … turned out more complicated than I thought. :frown:

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

Now this will work:

in your
ehrenfest said:
[tex] 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 [/tex]

[tex] 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 [/tex]

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 = … ? :smile:
 

1. What are zeros of a polynomial?

Zeros of a polynomial are the values of the variable that make the polynomial equation equal to zero. They are also known as roots or solutions of the polynomial.

2. How do you find the zeros of a polynomial?

To find the zeros of a polynomial, you can use the factor theorem, which states that if a value of the variable x makes the polynomial equal to zero, then x-a is a factor of the polynomial. You can also use the quadratic formula or long division to find the zeros.

3. Can a polynomial have more than one zero?

Yes, a polynomial can have multiple zeros. In fact, the number of zeros of a polynomial is equal to its degree. For example, a quadratic polynomial can have a maximum of two zeros, while a cubic polynomial can have up to three zeros.

4. What is the difference between real and complex zeros of a polynomial?

Real zeros of a polynomial are values of the variable that are real numbers, while complex zeros are values that involve imaginary numbers. Real zeros can be found by solving the polynomial equation, while complex zeros require the use of the quadratic formula.

5. How are zeros of a polynomial related to the graph of the polynomial?

The zeros of a polynomial are the x-intercepts of its graph. This means that the x-values at which the polynomial crosses the x-axis are the zeros of the polynomial. The graph of a polynomial can also help visualize the number of zeros and their approximate values.

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