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anonymous

  • 5 years ago

how do I expand and simplify the first 3 terms of: (3y + 5)^9 and (3b^2 - 2/b)^14?

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  1. King
    • 5 years ago
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    for 3y+5)^9 use pascals triangle

  2. anonymous
    • 5 years ago
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    it's still the same thing? But I'm going to be using pascal's triangle for both of these binomial powers..

  3. King
    • 5 years ago
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    ya fr both...

  4. anonymous
    • 5 years ago
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    but I kind of don't get (3b^2 - 2/b)^14.. o.o

  5. King
    • 5 years ago
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    take 1 as x and the other as y and substitute

  6. anonymous
    • 5 years ago
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    huh?

  7. anonymous
    • 5 years ago
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    Instead of using Pascal's triangle, you can use your scientific calculator. Look for the function that has "nCr" or "xCy" on it, or anything of the like that has the "C" in the middle. And then, with that, you can use the binomial expansion formula. Let's take this as an example: \[(a+b)^5 = a^5 + 5C1(a^4)(b^1) + 5C2(a^3)(b^2) + 5C3(a^2)(b^3)+5C4(a^1)(b^4)+b^5\] Wherein 5C1 is what you press on your calculator. Basically, the "a's" power decreases, and the b's powers increases. That's the easy way to put it.

  8. anonymous
    • 5 years ago
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    where do I find that button? o.O

  9. anonymous
    • 5 years ago
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    \[ (3y + 5)^9 = (3y)^9 + 9(3y)^8(5)^1+36(3y)^7(5)^2+84(3y)^6(5)^3...\] Do you have a scientific calculator?

  10. anonymous
    • 5 years ago
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    yes, I do.

  11. anonymous
    • 5 years ago
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    isn't y supposed increasing not decreasing?

  12. anonymous
    • 5 years ago
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    Hold up, let me get my textbook. I just learned this stuff today, too.

  13. anonymous
    • 5 years ago
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    oh xD

  14. anonymous
    • 5 years ago
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    Number on the left's power decreases, number on right's power increases.

  15. anonymous
    • 5 years ago
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    you don't need a calculator for first 3 terms \[\dbinom{9}{0}=1\] \[\dbinom{9}{1}=9\] \[\dbinom{9}{2}=\frac{9\times 8}{2}=9\times 4=36\] so you know the first three coefficients will be 1,9,36 then \[1\times (3y)^9+9\times (3y)^8\times 5^1+36\times (3y)^7\times 5^2\] are the first three terms, clean up with some arithmetic

  16. anonymous
    • 5 years ago
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    Oh. OH. I misread the question. The first three terms... right.

  17. anonymous
    • 5 years ago
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    for the exponent of 14 it is \[\dbinom{14}{0}=1,\dbinom{14}{1}=14, \dbinom{14}{2}=\frac{14\times 13}{2}=7\times 13=91\] only the last one requires any real computation

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