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DeadShot

  • one year ago

Determine the zeros of f(x) = x4 - x3 + 7x2 - 9x - 18.

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  1. shubhamsrg
    • one year ago
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    -1 fits in.

  2. shubhamsrg
    • one year ago
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    Got it, break 7 as 9-2 /

  3. DeadShot
    • one year ago
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    so \[\pm 1\] is the answer?

  4. shubhamsrg
    • one year ago
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    nopes.

  5. shubhamsrg
    • one year ago
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    What do get after writing 7 as 9-2 ?

  6. DeadShot
    • one year ago
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    \[x^4 - x^3 + 9 - 2x^2 - 9x -18\] ?

  7. shubhamsrg
    • one year ago
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    x^4 - x^3 + 9x^2 -2x^2 - 9x - 18.=0 =>(x^4 + 9x^2) - (x^3 -9x) -(2x^2 -18) =0 You see what should be the next step ?

  8. DeadShot
    • one year ago
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    yeah, factor each binomial, right?

  9. shubhamsrg
    • one year ago
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    yep.

  10. DeadShot
    • one year ago
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    (x^2 + 1) (x^2 + 9) for the first one?

  11. shubhamsrg
    • one year ago
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    try again.

  12. DeadShot
    • one year ago
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    I'm confused

  13. amoodarya
    • one year ago
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    x1 = 2 x2 = -1 x3 = 0 + 3i x4 = 0 - 3i

  14. DeadShot
    • one year ago
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    So, substitute (0 - 3i) and -1 for x^2, making it ((0 - 3i) + 9(-1))?

  15. DeadShot
    • one year ago
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    so, then it would become (-3i - 9)

  16. amoodarya
    • one year ago
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    hint : divide f(x) to x^2+9 you would find your answer!

  17. DeadShot
    • one year ago
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    So, divide x^4 - x^3 + 7x^2 - 9x - 18 by x^2 + 9 ?

  18. DeadShot
    • one year ago
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    so it would be x^2 - x + 7 - x - 2 ?

  19. DeadShot
    • one year ago
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    combine like terms, and it would be x^2 - 2x + 5 , right?

  20. zordoloom
    • one year ago
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    Hmm.. Looks like you have 2 real zeros and 2 imaginary.

  21. zordoloom
    • one year ago
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    Do you still need help?

  22. DeadShot
    • one year ago
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    yes, I don't know what to do next

  23. DeadShot
    • one year ago
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    I got it! Thanks!

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