Question

PLEASE HELP OR I FAIL IB MATH PLEASE in the expansion (3x-2)^12 the term in x^5 can be expressed as (12 choose r)*(3x)^p*(-2)^q

Answer 1

`C^(12)_7(3x)^5(-2)^7`

Explanation:

Use binomial expansion formula:

`(a+b)^n=\sum \limits _(k=0)^nC^n_ka^(n-k)b^(k)`

Then

`(3x-2)^(12)=\sum \limits_(k=0)^(12)C_k^(12)(3x)^(12-k)(-2)^k`

In the expansion
`(3x-2)^(12),` the term in
`x^5` is determined for

`12-k=5\\ \\-k=5-12\\ \\-k=-7\\ \\k=7,`

then the coefficient at
`x^5` is

`C^(12)_7(3x)^(12-7)(-2)^7=C^(12)_7(3x)^5(-2)^7`