Results 1 to 8 of 8

Math Help - A 3th degree polynome problem

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    6

    A 3th degree polynome problem

    hello I don't know how to solve a problem:

    P= x^3+x^2+Ax+1

    at A the value of P is a rational number for any x that satisfies the equation x^2+2x-2=0, and in this case P=B

    They ask me to find A and B

    I do have the answer, but what I need is to show me the method.

    Thank you very much!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,449
    Thanks
    1864
    Quote Originally Posted by Viou View Post
    hello I don't know how to solve a problem:

    P= x^3+x^2+Ax+1

    at A the value of P is a rational number for any x that satisfies the equation x^2+2x-2=0, and in this case P=B

    They ask me to find A and B

    I do have the answer, but what I need is to show me the method.

    Thank you very much!
    x^2+ 2x- 2= x^2+ 2x+ 1- 3= (x+1)^2- 3= 0
    so "any x that satisfies the equation x^2+ 2x- 2= 0" simply means x= -1+\sqrt{3} and -1-\sqrt{3}.

    If you put x= -1+ \sqrt{3} into the formula x^3+ x^2+ Ax+ 1 you get [tex](-1+\sqrt{3})^3+ (-1+ \sqrt{3})^2+ A(-1+\sqrt{3})+ 1= 6\sqrt{3}- 10+ 4- 2\sqrt{3}- A- A\sqrt{3}+ 1= -5- A+ (4+A)\sqrt{3}[/itex].

    Do you see what A must be in order that P be a rational number?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    6
    Quote Originally Posted by HallsofIvy View Post
    x^2+ 2x- 2= x^2+ 2x+ 1- 3= (x+1)^2- 3= 0
    so "any x that satisfies the equation x^2+ 2x- 2= 0" simply means x= -1+\sqrt{3} and -1-\sqrt{3}.

    If you put x= -1+ \sqrt{3} into the formula x^3+ x^2+ Ax+ 1 you get (-1+\sqrt{3})^3+ (-1+ \sqrt{3})^2+ A(-1+\sqrt{3})+ 1= 6\sqrt{3}- 10+ 4- 2\sqrt{3}- A- A\sqrt{3}+ 1= -5- A+ (4+A)\sqrt{3}.

    Do you see what A must be in order that P be a rational number?
    No I don't. What you did is exactly what I did before. The problem is what should I do from here.
    After I found that f(  -1+ \sqrt{3}) = -5 - A +(4+A)\sqrt{3}, what should I do?

    Wh
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    If x^2+2x-2=0 then x^2 = -2x+2. Also (multiply by x) x^3 = -2x^2+2x = -2(-2x+2)+2x = 6x-4. Therefore

    P= x^3+x^2+Ax+1 = (6x-4) + (-2x+2) + Ax + 1 = (A+4)x -1.

    Now you can put x = -1+\sqrt3, which is much easier than doing so at an earlier stage.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    6
    Quote Originally Posted by Opalg View Post
    If x^2+2x-2=0 then x^2 = -2x+2. Also (multiply by x) x^3 = -2x^2+2x = -2(-2x+2)+2x = 6x-4. Therefore

    P= x^3+x^2+Ax+1 = (6x-4) + (-2x+2) + Ax + 1 = (A+4)x -1.

    Now you can put x = -1+\sqrt3, which is much easier than doing so at an earlier stage.

    Thanks, Opalg, that really makes the counting easier.
    But it doesn't solve the problem, we still have the unknown A. Can you tell me what should I do next? How can I find A , knowing that
    f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Viou View Post
    Thanks, Opalg, that really makes the counting easier.
    But it doesn't solve the problem, we still have the unknown A. Can you tell me what should I do next? How can I find A , knowing that
    f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3
    P = B for this value of x. That is one equation in terms of A and B. Now use your other x value. That will get you a second equation in terms of A and B...

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Viou View Post
    How can I find A , knowing that
    f( \sqrt3 - 1)= - A - 5 + (A+4)\sqrt3
    You are told that P is rational, so you must ensure that the coefficient of \sqrt3 is zero.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2011
    Posts
    6
    Quote Originally Posted by topsquark View Post
    P = B for this value of x. That is one equation in terms of A and B. Now use your other x value. That will get you a second equation in terms of A and B...

    -Dan

    I did so before, but what I got is only the statement that 4\sqrt(3)= -4\sqrt(3) so its not very helpful

    Quote Originally Posted by Opalg View Post
    You are told that P is rational, so you must ensure that the coefficient of \sqrt3 is zero.
    Oh thank you, now I get it!!!!!

    B being a rational number AND
    f( \sqrt3 - 1)= B
    f( \sqrt3 - 1) = - A - 5 + (A+4)\sqrt3

    (A+4)\sqrt3=0
    A+4 = 0
    A = -4
    B =  - A - 5
    B = -(-4) - 5
    B = -1

    yay! happy ^w^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. second order first degree linear ODE problem
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 10th 2011, 03:01 PM
  2. Replies: 4
    Last Post: April 7th 2011, 11:08 AM
  3. Degree Problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 8th 2009, 11:51 PM
  4. Algebrah + Polynome, need help!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 11th 2008, 02:56 PM
  5. degree problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 16th 2005, 02:03 AM

/mathhelpforum @mathhelpforum