Question

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

Answer 1

opinion ?

Answer 2

what sign the exponent 5 ?

Answer 3

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

Answer 4

for you can solve it easy

Answer 5

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 ?

Answer 6

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 ?

Answer 7

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

Answer 8

It will be \[a3^0(2y)^5+ba3^1(2y)^4+ca3^2(2y)^3+da3^3(2y)^2+ea3^4(2y)^1+fa3^5(2y)^0\]

Answer 9

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.

Answer 10

http://www.wolframalpha.com/input/?i=expand+%283%2B2y%29^5

Answer 11

i just got it myself. BTW thanks all of you.