Results 1 to 5 of 5

Math Help - I'm having a hard time factoring this

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    5

    I'm having a hard time factoring this

    Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this. btw the original function is x*(x+1)^2

    I've asked a tutor at my school's tutoring center and she said that I should use my calc to find the zeros, which visually look like 1 and ~-2.5.


    Any help would be much appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by robo_robb View Post
    Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this.
    Rational Root Theorem

    Synthetic Division
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    5
    That certainly rings a bell. Thank you very much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by robo_robb View Post
    Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this. btw the original function is x*(x+1)^2

    I've asked a tutor at my school's tutoring center and she said that I should use my calc to find the zeros, which visually look like 1 and ~-2.5.


    Any help would be much appreciated!
    y = x (x + 1)^2 = x^3 + 2x^2 + x \Rightarrow \frac{dy}{dx} = 3x^2 + 4x + 1. There should be no trouble solving \frac{dy}{dx} = 3x^2 + 4x + 1 = 0.

    I don't see where y = 4x^3 + 9x^2 + 6x + 1 fits in .... is this another function for which you have to find the zeros of its derivative? Again, the derivative is a quadratic. and there is no trouble.

    Or are you trying to solve 0 = 4x^3 + 9x^2 + 6x + 1 in which case you note that by inspection x = -1 is a solution. Therefore x + 1 is a factor. Divide the factor into the cubic to get (x + 1)(4x^2 + 5x + 1) = 0. Now solve 4x^2 + 5x + 1 = 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    5
    Oh I'm sorry, I meant to say x*(x+1)^3

    All of the methods posted thus far are a great help. Thanks everyone.

    My roots turned out to be -1, -(1/4)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard factoring problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 12th 2011, 12:41 PM
  2. Replies: 6
    Last Post: August 23rd 2011, 01:22 PM
  3. Hard Time Applying a Boundary Condition
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 11th 2011, 06:07 PM
  4. Two integrals giving me a hard time
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 16th 2009, 02:24 PM
  5. hard time with graph problems
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 12th 2007, 09:44 AM

Search Tags


/mathhelpforum @mathhelpforum