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anonymous

  • one year ago

(x-(5+8i))(x+(5+8i))

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  1. rishavraj
    • one year ago
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    (a - b)(a + b) = a^2 - b^2

  2. anonymous
    • one year ago
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    that doesnt help me though because i already knew that but this equation is far more difficult and i's are involved @rishavraj

  3. anonymous
    • one year ago
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    i need steps

  4. rishavraj
    • one year ago
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    see i^2 = -1

  5. phi
    • one year ago
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    use FOIL on (5+8i)(5+8i)

  6. anonymous
    • one year ago
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    80i-39?

  7. anonymous
    • one year ago
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    then what

  8. phi
    • one year ago
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    yes, but we usually put the "real" component first (tradition) -39+80i

  9. phi
    • one year ago
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    that is the b^2 in a^2 - b^2 a^2 is x^2 so the answer is x^2 -( -39+80i) or x^2 +39 -80 i

  10. anonymous
    • one year ago
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    okay thank you so much

  11. rhr12
    • one year ago
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    http://www.tiger-algebra.com/drill/(x-(5_8i)(x_(5_8i)/

  12. anonymous
    • one year ago
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    i also have to multiply it by (x-4) and (x+14) or (x^2+10x-56) how would i do that? @phi

  13. anonymous
    • one year ago
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    because when i do it i get terms with both an i and an x and i dont know how to simplify that

  14. phi
    • one year ago
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    painfully. you want to do (x^2 + (39 -80 i)) (x^2+10x-56) ?

  15. anonymous
    • one year ago
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    yeah haha i tried and it just gets too confusing after awhile

  16. phi
    • one year ago
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    if we write it this way ,so we have "real" and imaginary terms: \[( (x^2+39) - 80i) (x^2+10x-56) \] and distribute the (x^2+10x-56) we get \[ (x^2+39) (x^2+10x-56) + -80(x^2+10x-56) i \] the first part gives a 4th order polynomial the second part (with the "i") will be the imaginary part. (we leave the "i" on the outside)

  17. phi
    • one year ago
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    no matter how you write it, it will be an ugly expression.

  18. anonymous
    • one year ago
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    oh i didnt know you could do that. so should i distribute it all out?

  19. anonymous
    • one year ago
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    i dont even know how to distribute the imaginary part

  20. anonymous
    • one year ago
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    for the first part i got \[x ^{4}+10x ^{3}-17x ^{2}+390x-2184\]

  21. phi
    • one year ago
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    for the imaginary part, you could distribute the -80 to get (-80x^2-800x+4480) i but multiplying it out really depends on what you plan to do next. Personally I would leave things factored... unless there is a reason to multiply things out.

  22. anonymous
    • one year ago
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    i would leave it factored also but the question was this: Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 4, -14, and 5 + 8i

  23. phi
    • one year ago
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    oh. In that case we want *** real coefficients **** and that means its roots come in complex conjugates in other words (oh my!) we should start 5+8i and 5-8i (notice the -8i)

  24. phi
    • one year ago
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    thus we would do (x - (5+8i))(x - (5 -8i))

  25. anonymous
    • one year ago
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    wow i just realized that okay thank you

  26. phi
    • one year ago
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    or, collecting the real part \[ ( (x-5) -8i) ( (x-5) + 8i) \] the answer is a^2 - b^2 where a is (x-5) and b is 8i

  27. anonymous
    • one year ago
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    so should i foil x-5 first?

  28. phi
    • one year ago
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    yes, FOIL (x-5)(x-5) you get \[ (x-5)^2 - (64 i^2) \\ x^2 -10x +25 + 64 \] notice the very convenient fact, the imaginary part disappears

  29. anonymous
    • one year ago
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    yes thank god then what about the other roots 4 and -14? do i just multiply (x-4) and (x+14) into that?

  30. phi
    • one year ago
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    yes

  31. anonymous
    • one year ago
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    okay thank youuuuuu ill tell you what i get

  32. phi
    • one year ago
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    \[ ( x^2 -10x +89)(x-4)(x+14) \]

  33. anonymous
    • one year ago
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    okay i got \[x ^{4}-67x ^{2}+1450x-4984\] sorry it took so long @phi

  34. phi
    • one year ago
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    \[x^4-67 x^2+1450 x-4984\] looks good

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