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.

All you need of the expansion of $(2-x)^6$ is the first three terms, $$(2-x)^6=2^6-(6)2^5x+(15)2^4x^2+\cdots=64-192x+240x^2+\cdots$$ Then multiplying by $1+2x+ax^2$ you can pick out the coefficient of $x^2$ as $$(1)(240)-(2)(192)+64a$$
