Results 1 to 6 of 6

Math Help - Please help me with 4 quick polynomials

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    2

    Please help me with 4 quick polynomials

    I just can't figure out these 4...Would be very happy if someone could help me out
    Solve the equations:
    1) (2x^3 -3x^2 -3x+2) / (x+1) < 0

    2) (x^3 -2x^2 -3x) > 0

    Solve the divisions
    3) (2x^3 +9x^2 -7x -4) : (x^2 +2x -3)

    4) x^4 : (x^2 - 4)
    Edit/Delete Message
    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 pizdac
    I just can't figure out these 4...Would be very happy if someone could help me out
    Solve the equations:
    1) (2x^3 -3x^2 -3x+2) / (x+1) < 0
    Do the division

    f(x)=\frac{2x^3-3x^2-3x+2}{x+1}=2x^2-5x+2.

    Now as the coefficient of x^2 is positive f(x)<0 between the roots
    of the quadratic on the right (if these are real). The roots are 2 and 1/2. So:

    f(x)=\frac{2x^3-3x^2-3x+2}{x+1}<0, when x \epsilon (1/2,2)

    RonL
    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 pizdac
    2) (x^3 -2x^2 -3x) > 0
    The LHS of this inequality is a cubic, it has either three real roots, or but
    a single root. Also as x \rightarrow \infty the cubic is positive.

    So if there is only a single root the cubic is >0 to the right
    of the root.

    If there are three roots it is >0 to the right of the greatest root,
    and again between the other two roots.

    (Sketch a cubic and you will see why this is so).

    As there is no constant term in the cubic we can factor it:

    x(x^2 -2x -3),

    and so we can find the other roots using the quadratic formula.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Problem 4
    \frac{x^4}{x^2-4}=x^2+4+\frac{16}{x^2-4}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Problem 2)
    x^3-2x^2-3x>0
    Factor,
    x(x^2-2x-3)
    Once more,
    x(x-3)(x+1).
    Now find the zero,
    x=-1,0,3
    Now observe each interval.
    x<-1
    -1<x<0
    0<x<3
    3<x
    Just pick points and see if the polynomial is positive or negative.
    Doing that we find that
    x<-1 This is Negative (Wrong)
    -1<x<0 This is Positive (Correct)
    0<x<3 This is Negative (Wrong)
    3<x This is Positive (Correct)
    Answer:
    -1<x<0 \mbox{ or } x>3
    Or another way of writing x\in (-1,0)\cup (3,\infty).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2006
    Posts
    2
    Thank you very much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quick question about cubing polynomials.
    Posted in the Algebra Forum
    Replies: 6
    Last Post: June 1st 2011, 10:04 AM
  2. [SOLVED] quick question on rings of polynomials.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 4th 2011, 10:03 AM
  3. Three quick questions - Vector spaces and polynomials
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 27th 2010, 06:08 AM
  4. Replies: 7
    Last Post: January 8th 2010, 03:13 AM
  5. Replies: 4
    Last Post: October 9th 2008, 12:51 PM

Search Tags


/mathhelpforum @mathhelpforum