degree 5, zero -3, multiplicity 2, zero -1, multiplicity 1, zero 2i, multiplicity 1. ?

build a polynomial function

4 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    Is this supposed to make sense?

    .

  • 1 decade ago

    I'm guessing you want to find a polynomial that has those roots. If so:

    List the following roots one time for each multiplicity, and make a factor from them:

    x = -3 <--> x + 3 = 0

    x = -3 <--> x + 3 = 0

    x = -1 <--> x + 1 = 0

    x = 2i <--> x - 2i = 0

    Since all the coefficients are real, the roots must come in complex conjugates, so:

    x = -2i <--> x + 2i = 0

    Since you have all 5 factors, just simply multiply the factors together:

    (x + 3)(x + 3)(x + 1)(x - 2i)(x + 2i)

    (x^2 + 6x + 9)(x + 1)(x^2 + 4)

    (x^3 + 7x^2 + 15x + 9)(x^2 + 4)

    x^5 + 7x^4 + 19x^3 + 37x^2 + 60x + 36

  • 4 years ago

    i assume the two have multiplicity 2? properly the real aspects are common, specifically that (x-5)(x-5) = x^2 -10x + 25 is a polynomial with roots 5, multiplicity 2. Now you may no longer have 3+2i happen with multiplicity 2, as this is conjugate must additionally be a root. yet you will discover that the polynomial that has 3+/- 2i as roots is x^2-6x+13 subsequently your polynomial is (x^2-10x +25)(x^2-6x+13) in case you're thinking a thank you to get the 2nd polynomial, this is common. you comprehend that the inspiration would desire to be x = 3+2i, so do here math to get a polynomial (x-3) = 2i (x-3)² = -4 x^2 - 6x +9 = -4 x² - 6x + 13= 0

  • Anonymous
    1 decade ago

    Why oh WHY don't people just use standard, normal everyday language and notation.

    I read.....and re-read your question. I have no idea what you're talking about, and from the looks of it, I'm not the only one.

Still have questions? Get your answers by asking now.