Results 1 to 7 of 7

Math Help - Finding the k value of a cubic equation with factor theorem

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    45

    Finding the k value of a cubic equation with factor theorem

    Hi, I'm working with the factor theorem and need help with the following question:

    If x^3 + 4x^2 + kx - 5 is divisible by (x+2), what is the value of k?

    I thought of doing the following

    f(-2)= (-2)^3 + 4(-2)^2 -2k - 5, and then replacing f(-2) with 0 since if f(x) is evenly divisible by d(x), then f(p) should equal 0, but obviously my thinking can't be correct since that would mean k = 0. I would be extremely grateful for any help. Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jul 2007
    Posts
    90
    You could do synthetic division and put a k as the coefficient of x to arrive at -5-2(k-4)=0

    Since you know the remainder should be 0. Now you have:
    -5-2(k-4)=0 \Longleftrightarrow -5-2k+8=0 \Longleftrightarrow 3-2k=0 \Longleftrightarrow k=\frac{3}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2007
    Posts
    45
    Quote Originally Posted by rualin View Post
    You could do synthetic division and put a k as the coefficient of x to arrive at -5-2(k-4)=0

    Since you know the remainder should be 0. Now you have:
    -5-2(k-4)=0 \Longleftrightarrow -5-2k+8=0 \Longleftrightarrow 3-2k=0 \Longleftrightarrow k=\frac{3}{2}
    Thanks so much for the help, but I've actually never done/been taught synthetic division. Is there another method I could use?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    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 starswept View Post
    Thanks so much for the help, but I've actually never done/been taught synthetic division. Is there another method I could use?
    you're original method was fine. you simply made an error in calculation somewhere

    (-2)^3 + 4(-2)^2 - 2k - 5 = 0

    \Rightarrow -8 + 16 - 2k - 5 = 0

    \Rightarrow 3 - 2k = 0

    \Rightarrow \boxed { k = \frac {3}{2} }

    Using synthetic division is fine, but it does not really rely on the factor theorem as this method does, so i believe this is the method you should use
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2007
    Posts
    90
    That's simpler and less work than through synthetic division.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2007
    Posts
    45
    Oh gosh, I'm sorry, that was a dumb mistake on my part. After 12 straight hours of math, apparently I've forgotten how to add and subtract. Following through on the original method is perfect. Thank you both for the help!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    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 starswept View Post
    After 12 straight hours of math, apparently I've forgotten how to add and subtract.
    I know how that feels, so let's all put that little mistake behind us
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] factor a cubic polynomial
    Posted in the Algebra Forum
    Replies: 6
    Last Post: January 10th 2012, 04:30 PM
  2. Limit with cubic factor
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: April 5th 2011, 06:03 AM
  3. Finding equation of a cubic function?
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: March 20th 2010, 02:59 PM
  4. Replies: 1
    Last Post: August 24th 2009, 02:30 AM
  5. Help with factor theorem and quintic equation
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: July 6th 2007, 07:24 PM

Search Tags


/mathhelpforum @mathhelpforum