Results 1 to 11 of 11

Math Help - Cubic Equations

  1. #1
    Senior Member
    Joined
    Jul 2007
    Posts
    290

    Cubic Equations

    I forgot how to do cubic equations please help


    2x^3 + x^2 - 18x - 9 = 0



    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by JonathanEyoon View Post
    I forgot how to do cubic equations please help


    2x^3 + x^2 - 18x - 9 = 0



    Thanks in advance
    Use the rational roots theorem to suggest factors, in this case (x+1/2) appears to be a factor, then divide through to get:

    2x^3 + x^2 - 18x - 9 = 2(x+1/2)(x^2-9) = 2(x+1/2)(x-3)(x+3)=0

    so the roots are x=-1/2, x=3, x=-3.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    I forgot how to do cubic equations please help


    2x^3 + x^2 - 18x - 9 = 0



    Thanks in advance
    this is a nice one, we can factor by groups.

    2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)

    now factor out the common (2x + 1), we get:

    (2x + 1) \left( x^2 - 9 \right) = 0

    can you take it from here?

    EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jhevon View Post
    this is a nice one, we can factor by groups.

    2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)

    now factor out the common (2x + 1), we get:

    (2x + 1) \left( x^2 - 9 \right) = 0

    can you take it from here?

    EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff
    Welcome back slow coach

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Jhevon View Post
    this is a nice one, we can factor by groups.

    2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)

    now factor out the common (2x + 1), we get:

    (2x + 1) \left( x^2 - 9 \right) = 0

    can you take it from here?

    EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff

    { -3 , -1/2 , 3} correct?

    Another question is, usually in a cubic equation, using the method you used is always gonna be the more common method to solving these right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    { -3 , -1/2 , 3} correct?
    yes (as CaptainBlack said)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Jhevon View Post
    yes (as CaptainBlack said)


    Thanks alot to the both of you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by JonathanEyoon View Post
    Another question is, usually in a cubic equation, using the method you used is always gonna be the more common method to solving these right?
    Most cubic equations cannot be solved using the methods that CaptainBlack
    and Jhevon used, though probably the majority that you will meet on low
    level HS and University maths courses will be. The usual way of solving them
    is numerically, and if we are feeling adventurous using Ferro-Tartaglia-Cardano method.

    Note a cubic with real coefficients always has at least one real root/factor

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jul 2007
    Posts
    90
    Quote Originally Posted by CaptainBlack View Post
    Most cubic equations cannot be solved using the methods that CaptainBlack
    and Jhevon used, though probably the majority that you will meet on low
    level HS and University maths courses will be. The usual way of solving them
    is numerically, and if we are feeling adventurous using Ferro-Tartaglia-Cardano method.

    Note a cubic with real coefficients always has at least one real root/factor

    RonL
    Must cubics with real coefficients always have an odd number of real roots? If so, is it generally true that for all polynomials of odd degree with real coefficients there are an odd number of real roots and for those of even degree an even number of real roots?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,067
    Thanks
    370
    Awards
    1
    Quote Originally Posted by rualin View Post
    Must cubics with real coefficients always have an odd number of real roots? If so, is it generally true that for all polynomials of odd degree with real coefficients there are an odd number of real roots and for those of even degree an even number of real roots?
    Yes. The reason is that for a polynomial with real coefficients any complex roots show up as complex conjugates. So the complex roots always come in pairs. (I have forgotten the name for this theorem, sorry!)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by topsquark View Post
    Yes. The reason is that for a polynomial with real coefficients any complex roots show up as complex conjugates. So the complex roots always come in pairs. (I have forgotten the name for this theorem, sorry!)

    -Dan
    Like halfwits

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubic Equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 10th 2011, 06:36 AM
  2. Cubic equations!
    Posted in the Algebra Forum
    Replies: 5
    Last Post: June 3rd 2011, 03:29 AM
  3. Solve 2 cubic equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 3rd 2010, 11:30 PM
  4. Solving Cubic Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 18th 2009, 12:32 AM
  5. Cubic equations and Matlab
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2007, 01:12 PM

Search Tags


/mathhelpforum @mathhelpforum