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? 發問於 Science & MathematicsMathematics · 6 年前

How do I find the zeros for a polynomial?

How do I find the zeros for the polynomial x^4-10x^3+23x^2-10x+22?

4 個解答

相關度
  • 最愛解答

    First off, I notice that the 2nd and 4th coefficients are the same

    Then I noticed that the 3rd coefficient is the sum of the first and last coefficient

    This leads me to believe that I can factor this by grouping

    x^4 - 10x^3 + 22x^2 + x^2 - 10x + 22 =>

    x^2 * (x^2 - 10x + 22) + 1 * (x^2 - 10x + 22) =>

    (x^2 + 1) * (x^2 - 10x + 22)

    Now, x^2 + 1 is not factorable. Let's look at x^2 - 10x + 22

    x^2 - 10x + 22 = 0

    x^2 - 10x + 25 = 3

    (x - 5)^2 = 3

    x - 5 = +/- sqrt(3)

    So that's not factorable either

    (x^2 + 1) * (x^2 - 10x + 22)

    That's it.

    The zeros are -i , i , 5 + sqrt(3) , 5 - sqrt(3)

  • 匿名
    6 年前

    The Captain's answer is brilliant. He sees thing there that I can't.

    So here's a different approach.

    First check for rational roots, which the rational root test tells you must be ±1, ±2, ±11 or ±22. Unfortunately, none of those are roots.

    Perhaps you can factor it as (x^2 + ax + b)(x^2 + cx + d)

    = x^4 + (a+c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.

    Now you equate the four coefficients to what you have to start with, and solve for the four unknowns, a, b, c, and d.

    You have

    a+c=-10

    b+d+ac=23

    ad+bc=-10

    bd=22

    Since we want these to all be integers, there are limited possibilities for b and d. If we assume that b and d are 1 and 22, we can WLOG assign b to 1 and d to 22. That gives us:

    b+d+ac=23 so ac= 0.

    a+c=-10 so a and b must be 0 and -10 (not sure yet which is which).

    ad+bd=-1 ==> 22a + 1c = -10, so a=0 and c = -10

    So we are done: the factors are (x^2 + 1)(x^2 - 10x + 22). Now you can get the roots in the usual way.

    If you take b=2 and d=11, then you have 11a+2b = -10, and you can't solve that with integers a and b such that ab=-10

    There are similar ways to eliminate other possible ways of factoring 22 and ac, but I won't bother to go through them since we already have the answer. We also know that there is only one solution, and we've already found it. It's not like we are going to find another one. There's only one solution since Z[x], the ring of integer coefficient polynomials, is a Unique Factorization Domain.

    Not the most elegant solution, but it does eventually get to the answer.

  • 6 年前

    x = 5 - sqrt(3) AND 5 + sqrt(3).

    These are the "Real roots"

    x = i, - i

    These are the "Complex roots"

    You'll have to use a numerical method. Newton's method for instance could be used to find the real roots. https://en.wikipedia.org/wiki/Newton's_method

    Som

  • 6 年前

    Do synthetic division. I don't feel like explaining it because it's summer.

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