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anonymous

  • 5 years ago

Which polynomial has roots at... (see attachment)

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  1. anonymous
    • 5 years ago
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    A. 11x^2-8x+16 B. 11x^2-12x+12 C. 9x^2-12x+16 D. 9X^2-8x+12

  2. anonymous
    • 5 years ago
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  3. anonymous
    • 5 years ago
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    It's the second attach. sorry..

  4. anonymous
    • 5 years ago
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    it is \[9x^2-12x+16\] if you want i will show you how to get it without solving each one.

  5. anonymous
    • 5 years ago
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    sure. i never got up to learning that in alg 2. so it should be good to learn

  6. anonymous
    • 5 years ago
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    ok we start slow, but if it is too slow let me know. we know the zeros are \[\frac{2+2\sqrt{3}i}{3}\] and \[\frac{2+2\sqrt{3}i}{3}\] and we know that if a quadratic has zeros \[r_1\] and \[r_2\] it factors as \[(x-r_1)(x-r_2)\]

  7. anonymous
    • 5 years ago
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    so we can just write \[(x-\frac{2+\sqrt{3}i}{3})(x-\frac{2-\sqrt{3}i}{3})\]

  8. anonymous
    • 5 years ago
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    Okay

  9. anonymous
    • 5 years ago
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    get rid of the 3 in the denominator by multiplying each factor by 3 to get \[(3x-(2+\sqrt{3}i)(3x-(2-\sqrt{3}i)\]

  10. anonymous
    • 5 years ago
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    okay

  11. anonymous
    • 5 years ago
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    now this might look like a pain to multiply out but it is not. do not distribute to get 3 terms , just use the old 'first outer inner last"

  12. anonymous
    • 5 years ago
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    "first" is \[9x^2\]

  13. anonymous
    • 5 years ago
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    "last" is \[(2+2\sqrt{3}i)(2-\sqrt{3}i)\]

  14. anonymous
    • 5 years ago
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    yeah

  15. anonymous
    • 5 years ago
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    oops i meant \[(2+2\sqrt{3}i)(2-2\sqrt{3}i)\]

  16. anonymous
    • 5 years ago
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    because of + and - right?

  17. anonymous
    • 5 years ago
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    and when you multiply a complex number by its conjugate you get \[a^2+b^2\] so in this case you get \[2^2+(2\sqrt{3})^2=4+12=16\]

  18. anonymous
    • 5 years ago
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    yes because the zeros are \[a+bi\] and \[a-bi\]

  19. anonymous
    • 5 years ago
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    ok

  20. anonymous
    • 5 years ago
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    so we know first term is \[9x^2\] and last term is \[16\] hope it is clear how i got 16 another example would be \[(3+\sqrt{5}i)(3-\sqrt{5}i)=3^2+5^2=34\]

  21. anonymous
    • 5 years ago
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    yeah okay

  22. anonymous
    • 5 years ago
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    ok now for the middle terms, the 'outer - inner' outer is \[3x(-2+\sqrt{3}i)\] inner is \[3x(-2-\sqrt{3}i)\] and when you add the imaginary part will cancel (add to zero, it always does!) and you just get \[-6x-6x=-12x\]

  23. anonymous
    • 5 years ago
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    Oh okay!

  24. anonymous
    • 5 years ago
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    Ooooh okay! Jeez, that was a lot. LOL. You figured it out quick. Are you a math teacher?

  25. anonymous
    • 5 years ago
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    may seem like a lot of work, but multiplying a complex number by its conjugate is easy, and seeing that in the middle terms the imaginary part drops out is not too hard. easier than solving 4 quadratics using the formula and risking an arithmetic error. good luck.

  26. anonymous
    • 5 years ago
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    Thanks!!

  27. anonymous
    • 5 years ago
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    welcome!

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