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anonymous

  • 5 years ago

. Write the complex binomial expansion. (5x+3)^4

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  1. dumbcow
    • 5 years ago
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    you have to know combinations like 4 choose1, 4 choose2

  2. amistre64
    • 5 years ago
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    14641 might work?

  3. anonymous
    • 5 years ago
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    i am so lost in this one....

  4. amistre64
    • 5 years ago
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    (5x)^4+4(5x)^3 (3)+6(5x)^2(3)^2+4(5x)(3)^3+(3)^4 if i see it right; but what do you refer to as complex?

  5. amistre64
    • 5 years ago
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    with imaginary numbers? or simply complicated?

  6. anonymous
    • 5 years ago
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    4C0 (5x)^4 + 4C1 (5x)^3 (3)^1 + 4C2 (5x)^2 (3)^2 + 4C3 (5x)^1 (3)^3 +4C4 (3^4)

  7. anonymous
    • 5 years ago
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    and im finished on here, going to go to sleep

  8. anonymous
    • 5 years ago
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    i dont know, thats what my problem says, exactly how i wrote it

  9. anonymous
    • 5 years ago
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    ^ "with imaginary numbers"? thats not even possible. They wanted the binomail expansion , they want the binomial coefficents, not just pascals triangle

  10. amistre64
    • 5 years ago
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    then elecs answer looks to be close to what they want :)

  11. anonymous
    • 5 years ago
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    so that long thing is the answer ? lol

  12. anonymous
    • 5 years ago
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    thats what they mean by "complex"

  13. anonymous
    • 5 years ago
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    4C0 (5x)^4 + 4C1 (5x)^3 (3)^1 + 4C2 (5x)^2 (3)^2 + 4C3 (5x)^1 (3)^3 +4C4 (3^4) This is the answer lol wow

  14. amistre64
    • 5 years ago
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    I had parts of it right lol ;)

  15. dumbcow
    • 5 years ago
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    you need to multiply it out it will simplify a little bit

  16. anonymous
    • 5 years ago
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    im still in shock that thats the long answer lol

  17. dumbcow
    • 5 years ago
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    yeah the bigger the exponent, the longer the answer

  18. amistre64
    • 5 years ago
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    where do we get the 4c1 and stuff from? the exponent?

  19. dumbcow
    • 5 years ago
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    yes

  20. amistre64
    • 5 years ago
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    :) dint know if that was a fluke, or not :)

  21. dumbcow
    • 5 years ago
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    no yours above was correct 4C1=4 4C2=6

  22. anonymous
    • 5 years ago
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    so elecengineer answer was correst right?

  23. amistre64
    • 5 years ago
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    ....i wanna say yes, but I dont have a combinatorics button on the calulator ... :/

  24. anonymous
    • 5 years ago
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    hmmm.... ok

  25. amistre64
    • 5 years ago
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    according to an online nCr calculator, mines good lol :)

  26. amistre64
    • 5 years ago
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    but you prolly want the 4C0, 4C1 stuff

  27. dumbcow
    • 5 years ago
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    4C0 = 1 4C1=4 4C2=6 4C3 = 4 4C4=1 they just represent numbers

  28. dumbcow
    • 5 years ago
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    nCr = n!/(n-r)!r!

  29. anonymous
    • 5 years ago
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    wow this is way over my head

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