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

1. jhonyy9

opinion ?

2. jhonyy9

what sign the exponent 5 ?

3. jhonyy9

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

4. jhonyy9

for you can solve it easy

5. jhonyy9

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

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

sorry there wann being (a+b)^2 = ???

8. henpen

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

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

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11. eliassaab