Given: the coefficent of $x^2$ in the expansion $(1+2x+ax^2)(2-x)^6$ 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?
ps. is there an easier method other than expanding the terms? I have tried using the bionomal expansion; however, one needs still to multiply the term. expand $(2-x)^6$ which is not very fast.