Results 1 to 4 of 4

Math Help - Finding Zeros

  1. #1
    Junior Member
    Joined
    Mar 2007
    Posts
    32

    Post Finding Zeros

    Hi, I m' not sure how to solve this.
    I'm asked to find the zeros using synthetic division for function
    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32

    I found that one of the zeros is 2 but I can't find the rest. Please help!
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Quote Originally Posted by cuteisa89 View Post
    Hi, I m' not sure how to solve this.
    I'm asked to find the zeros using synthetic division for function
    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32

    I found that one of the zeros is 2 but I can't find the rest. Please help!
    Thanks
    I love these kinds of problems:

    First you must understand that this problem requires tail over head.

    32=tail
    1=head

    First the factors of 32 the tail

    1,2,4,8,16,32 over the factors of the head which is one

    1

    Now you must put all of your factors together you solution can be + or -

    here we go:

    1
    2
    4
    8
    16
    32

    -1
    -2
    -4
    -8
    -16
    -32

    Which ones are the zeros?

    2

    This is your ONLY solution.

    I will let someone else show you how to use synthetic division because latex is down and it would be impossible for you understand where I am coming from.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by cuteisa89 View Post
    Hi, I m' not sure how to solve this.
    I'm asked to find the zeros using synthetic division for function
    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32

    I found that one of the zeros is 2 but I can't find the rest. Please help!
    Thanks
    As one of the zeros is 2 (x-2) is a factor of x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32,
    so divide the later by (x-2) to get the quartic that when multiplied by (x-2) will give
    you the quintic.

    After doing the division gives:

    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)(x^4 - 6x^3 + 16x^2 - 24x + 16)

    Now the rational root theorem suggests we try x=+/-1, +/-2, +/-4, +/-8, +/-16 as
    potential roots for the quartic, and we find x=2 is a root (in fact it is the only real root
    but I won't use this and will do this the long winded way)so dividing through again
    we find:

    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)^2 (x^3 - 4x^2 + 8x - 8)

    Now the rational root theorem suggests we try x=+/-1, +/-2, +/-4, +/-8 as potential
    roots for the cubic, and we find x=2 is a root so dividing through again we find:

    x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)^3 (x^2 - 2x + 4)

    Now the discriminant of the quadratc factor will tell us that x^2 - 2x + 4
    has no real roots, so the factorisation can proceed no further over the reals.

    So we conclude that the only real root is x=2.

    Of course we could have concluded earlier that there is a factor of the form:
    (x-2)^n, where n=1 or 3, and so saved ourselves some work.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the zeros
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 21st 2009, 06:18 PM
  2. Finding the zeros
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 16th 2009, 05:31 AM
  3. Help with Finding Zeros
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 12th 2008, 01:58 AM
  4. Finding the Zeros
    Posted in the Algebra Forum
    Replies: 10
    Last Post: March 20th 2007, 06:45 PM
  5. finding zeros
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 3rd 2007, 10:39 AM

Search Tags


/mathhelpforum @mathhelpforum