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muhammad9t5
 3 years ago
how to do this by binomial theorem (3+2y)^5
muhammad9t5
 3 years ago
how to do this by binomial theorem (3+2y)^5

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jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0what sign the exponent 5 ?

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0so if a^5 =a^2 *a^3 so than how can you writing your exercise ? fdor easy solve it

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0for you can solve it easy

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0let 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 ?

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 ?

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0sorry there wann being (a+b)^2 = ???

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0It will be \[a3^0(2y)^5+ba3^1(2y)^4+ca3^2(2y)^3+da3^3(2y)^2+ea3^4(2y)^1+fa3^5(2y)^0\]

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0Due to the symmetry of binomial expansion, a=f, b=e and c=d. To find what they ARE without bruteforcing your way expanding the brackets, look at the 5th (as it was ^5) row of Pascal's triangle.

muhammad9t5
 3 years ago
Best ResponseYou've already chosen the best response.0i just got it myself. BTW thanks all of you.
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