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muhammad9t5

  • 3 years ago

how to do this by binomial theorem (3+2y)^5

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  1. jhonyy9
    • 3 years ago
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    opinion ?

  2. jhonyy9
    • 3 years ago
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    what sign the exponent 5 ?

  3. jhonyy9
    • 3 years ago
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    so if a^5 =a^2 *a^3 so than how can you writing your exercise ? fdor easy solve it

  4. jhonyy9
    • 3 years ago
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    for you can solve it easy

  5. jhonyy9
    • 3 years ago
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    let 3 sign a and 2y sign b so than in this case you can writing that (a+b)^5 =(a+b)^2 *(a+b)^3 yes ?

  6. jhonyy9
    • 3 years ago
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    so than your exersice will be (3+2y)^5 =(3+2y)^2 *(3+2y)^3 so what you -hope so much that - can end it easy ,because you know the formula (a3b)^2 = ??? and formule for (a+b)^3 = ??? ok ? i think that this formules you have learned in your schooles till today ok ?

  7. jhonyy9
    • 3 years ago
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    sorry there wann being (a+b)^2 = ???

  8. henpen
    • 3 years ago
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    It will be \[a3^0(2y)^5+ba3^1(2y)^4+ca3^2(2y)^3+da3^3(2y)^2+ea3^4(2y)^1+fa3^5(2y)^0\]

  9. henpen
    • 3 years ago
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    Due to the symmetry of binomial expansion, a=f, b=e and c=d. To find what they ARE without brute-forcing your way expanding the brackets, look at the 5th (as it was ^5) row of Pascal's triangle.

  10. henpen
    • 3 years ago
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    |dw:1346412351008:dw|

  11. eliassaab
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=expand+%283%2B2y%29^5

  12. muhammad9t5
    • 3 years ago
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    i just got it myself. BTW thanks all of you.

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