solving for a coefficent term of factored polynomial. 
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$

I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4}-160\,a{x}^{5}+
60\,a{x}^{6}-12\,a{x}^{7}+a{x}^{8}
$
because of the given info $48x^2=64x^2-144x^2$ solve for $a,$ $a=3$.
Correct?
P.S. is there an easier method other than expanding the terms?
I have tried using the bionomal expansion; however, one needs still to multiply the terms.  Expand $(2-x)^6$ which is not very fast.

The coefficient of $x^2$ is obtained by adding three contributions :


*

*$a\, 2^6$ : the coefficient of $x^2$ at the left and $2\,x^0$ at the right ($2$ at power $6$)

*$2(6\cdot 2^5(-1))$: the $x$ coefficients on both sides ((constant term $2$)$^5$ $\times\ x$ coef. at the right)

*$1\binom{6}{2}2^4$ : the constant coefficient of the left $\times$ the $x^2$ coefficient at the right ($2^4$)


The sum of all this is : $\quad 2^6\, a-12\cdot 2^5+15\cdot 2^4=48$
Divided by $2^4=16$ this becomes $\ 4a-24+15=3\ $ i.e. $\ \boxed{a=3}$
