Results 1 to 4 of 4

Math Help - factoring sixth degrees

  1. #1
    Junior Member
    Joined
    Dec 2006
    Posts
    60

    factoring sixth degrees

    Can someone help me out here?

    x^6 - 7x^3 - 8 = 0

    THere must be a simpler way to factoring this without having to use factor theorem! Someone please help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    x^6-7x^3-8=0\implies(x^3+1)(x^3-8)=0

    Continue.
    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 checkmarks View Post
    Can someone help me out here?

    x^6 - 7x^3 - 8 = 0

    THere must be a simpler way to factoring this without having to use factor theorem! Someone please help?
    It's a quadratic in y=x^3

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2006
    Posts
    60
    ahh i thought of that.
    but i just assumed an exponent raised to an exponent, you would multiply...so (x^3)(x^3) = x^9
    shows you how terrible i am at simple math

    thanks guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Roots of a Sixth-Degree Polynomial
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 27th 2011, 02:53 AM
  2. Find six distanct sixth roots of -1 + i root 3
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 20th 2010, 03:11 AM
  3. Replies: 5
    Last Post: January 13th 2010, 11:50 AM
  4. Replies: 2
    Last Post: April 6th 2009, 09:19 AM
  5. HARD sixth grade problem
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: September 6th 2008, 09:54 AM

Search Tags


/mathhelpforum @mathhelpforum