Results 1 to 5 of 5

Math Help - polynomial

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    polynomial

    Given that 3+\sqrt{2} and 2-i are the roots of the equation f(x)=0 . If the degree of f is 4 and f(1)=8 . Find the polynomial f(x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    I suppose that f(x) is a polynomial with integer coefficients
    Therefore if 2-i is a solution then 2+i is a solution
    And if 3 - \sqrt{2} is a solution then 3 + \sqrt{2} is a solution
    Finally f(x) = \alpha (x-(2-i))(x-(2+i))(x-(3 - \sqrt{2})(x-(3 + \sqrt{2}))
    Expand and determine \alpha using the value of f(1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by running-gag View Post
    Hi

    I suppose that f(x) is a polynomial with integer coefficients
    Therefore if 2-i is a solution then 2+i is a solution
    And if 3 - \sqrt{2} is a solution then 3 + \sqrt{2} is a solution
    Finally f(x) = \alpha (x-(2-i))(x-(2+i))(x-(3 - \sqrt{2})(x-(3 + \sqrt{2}))
    Expand and determine \alpha using the value of f(1)
    THanks but i don understand how can u just reverse the operation sign and call it another root ? Is it because its a polynomial of degree 4 ?

    And where does the 'a' come from ?

    Thanks again ..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2009
    Posts
    66
    Quote Originally Posted by thereddevils View Post
    THanks but i don understand how can u just reverse the operation sign and call it another root ? Is it because its a polynomial of degree 4 ?

    And where does the 'a' come from ?

    Thanks again ..
    I can't quote a "rule" for Real roots, but I do know that if any roots of an equation are complex, they *always* come in pairs of conjugates. So if a +bi is a root, a - bi *has* to be a root as well.

    There's probably something similar for Real roots, but I do not know any formal properties in this area.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    When have you ever (and only) gotten solutions of these forms, with radicals or imaginaries (or both)? From the "plus-minus" part of the Quadratic Formula!

    Working backwards here from the zeroes, you know that, if one zero is "(something) plus (a square root)", then another zero must be "(that same thing) minus (that same square root)". Using the same reasoning with the complex root, you obtain the fourth zero.

    From these, and the fact that, if x = a is a root, then x - a is a factor, you can find all four factors, and multiply them out to find the original polynomial.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 23rd 2011, 07:36 AM
  2. Replies: 1
    Last Post: February 24th 2011, 07:46 PM
  3. Replies: 1
    Last Post: December 15th 2009, 08:26 AM
  4. [SOLVED] dividing polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 3rd 2008, 03:00 PM
  5. dividing a polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 2nd 2005, 01:26 AM

Search Tags


/mathhelpforum @mathhelpforum