Binomial expansion

[jess5085]

Hello,
I know how to do it without the (x+3) but I not sure how to do this question.

. . .Q. Find the coefficient of x⁵ in the expansion of (x + 3)(2x - 1)⁶.

Sure far I have attempted this but it doesn't seem to be working.

\(\displaystyle (x\, +\, 3)\,\binom{6}{1}\,(2x)^5\, (-1)^1\)

\(\displaystyle =\, (x\, +\, 3)\, (6)\, (32x^5)\, (-1)\)

\(\displaystyle =\, (x\, +\, 3)(-192x^5)\)

\(\displaystyle =\, -192x^6\, -\, (3)(192x^5)\)

Thanks in advance!

 

[daon2]

Hint:

\(\displaystyle (c_0x^0+\cdots +c_4x^4+c_5x^5+\cdots +c_6x^6)(x+3) = (c_0x+\cdots + \fbox{$c_4x^5$}+c_5x^6\cdots + c_6x^7) + (3c_0x^0+\cdots + 3c_4x^4+\fbox{$3c_5x^5$}\cdots + c_6x^6)\)