Results 1 to 3 of 3

Math Help - Finding real and complex zeros of x^5 + 8x^2

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    12

    Finding real and complex zeros of x^5 + 8x^2

    Let P(x) = x^5 + 8x^2. Find all the zeros of P, real and complex.

    I factor to this:
    0=x^2(x^3+8), giving me zeros of 0 and -2. ARE there any complex zeros? The reason I ask is because the next part of the question is "Express P as a product of linear and irreducible quadratic factors with real coefficients"

    Where am I going wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by cdbmath View Post
    Let P(x) = x^5 + 8x^2. Find all the zeros of P, real and complex.

    I factor to this:
    0=x^2(x^3+8), giving me zeros of 0 and -2. ARE there any complex zeros? The reason I ask is because the next part of the question is "Express P as a product of linear and irreducible quadratic factors with real coefficients"

    Where am I going wrong?
    There are complex zeros.

    Note that P(x)=x^2(x^3+8)={\color{blue}x^2(x+2)}{\color{red}  (x^2-2x+4)}

    The part highlighted in blue will give you the real solutions to P(x)=0 and the part highlighted in red will give you the complex solutions to P(x)=0.

    Also note that P(x)=x^2(x+2)(x^2-2x+4) is the answer to the second part of the question.

    Does this clarify things?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    12
    Quote Originally Posted by Chris L T521 View Post
    There are complex zeros.

    Note that P(x)=x^2(x^3+8)={\color{blue}x^2(x+2)}{\color{red}  (x^2-2x+4)}

    The part highlighted in blue will give you the real solutions to P(x)=0 and the part highlighted in red will give you the complex solutions to P(x)=0.

    Also note that P(x)=x^2(x+2)(x^2-2x+4) is the answer to the second part of the question.

    Does this clarify things?
    Thanks! I used long division to get (x^4-2x^3+4x^2) and then took out x^2 to get (x^2-2x+4)... I was then able to get my complex zeros.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding real numbers of a complex polynomial
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 13th 2011, 09:05 PM
  2. Finding All Real Zeros
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 10th 2009, 07:08 PM
  3. Replies: 5
    Last Post: May 6th 2009, 05:50 PM
  4. Finding all the possible real zeros
    Posted in the Algebra Forum
    Replies: 11
    Last Post: January 26th 2008, 02:06 PM
  5. Finding the real part of Complex Power
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 14th 2007, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum