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toadytica305

List all the real and imaginary zeros of f(x) = x^3+x^2+ 13x - 15... 1 is a zero of the function

  • one year ago
  • one year ago

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  1. satellite73
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    did you factor? you already knew 1 was a zero, from the last question

    • one year ago
  2. Callisto
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    I think the question asks to solve f(x) =0 Since the question gives you 1 is a zero of the function. (x-1) is a factor. Now, do the division to get the other factor first, can you?

    • one year ago
  3. satellite73
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    once you factor, you find the other zeros using the quadratic formula

    • one year ago
  4. toadytica305
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    You said (x−1)(x^2+2x+15) This gives zeros of 1, 3, 5, 15?

    • one year ago
  5. Callisto
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    Nope!~ f(x) =0 x^3+x^2+ 13x - 15 =0 (x−1)(x^2+2x+15) =0 So, x-1 =0 -(1) or (x^2+2x+15) =0 -(2) You can solve (1) easily. And solve (2) using quadratic formula!~

    • one year ago
  6. Callisto
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    For a quadratic equation ax^2 + bx + c=0 \[x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]^ Quadratic formula

    • one year ago
  7. toadytica305
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    i get \[(-4\sqrt{15}) /2\]

    • one year ago
  8. Callisto
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    It doesn't seems right... \[x=\frac{-2 \pm \sqrt{2^2 - 4(1)(15)}}{2(1)} =?\]

    • one year ago
  9. toadytica305
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    -2i√14?

    • one year ago
  10. Callisto
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    Nope.. \[x=\frac{-2\pm \sqrt {-56}}{2} = \frac{-2\pm -2\sqrt {-14}}{2} = -1 \pm -14i\]

    • one year ago
  11. satellite73
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    \[\sqrt{14}i\]

    • one year ago
  12. satellite73
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    a typo

    • one year ago
  13. Callisto
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    Sorry :S \[-1\pm -\sqrt{14}i\]

    • one year ago
  14. toadytica305
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    Im still not sure how I get the real and imaginary zeros off that...

    • one year ago
  15. Callisto
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    the real one is 1 the imaginary ones are \[-1\pm -\sqrt{14}i\] Actually, the question just asks to you to solve f(x) =0 ... I think..

    • one year ago
  16. toadytica305
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    Thankssss!

    • one year ago
  17. Callisto
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    Welcome!!~~~

    • one year ago
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