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.
 A: It would be much easier to just compute the coefficient at $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$. You can begin by computing:
$$ (2-x)^6 
= 64 - 6 \cdot 2^5 x + 15 \cdot 2^4 x^2 + x^3 \cdot (...) 
= 64 - 192 x + 240 x^2 + x^3 \cdot (...) $$
Now, multiply this by $(1+2x+ax^2)$. Again, you're only interested in the term at $x^2$, so you can spare yourself much effort by just computing this coefficient: to get $x^2$ in the product, you need to take 
$64, \ -192 x, \ 240 x^2$  from the first polynomial, and 
$ax^2,\ 2x, 1$ from the second one (respectively).
$$(1+2x+ax^2)(2-x)^6 = (...)\cdot 1 + (...) \cdot x 
+ (64a - 2\cdot 192 + 240 )\cdot x^2
+ x^3 \cdot(...) $$
Now, you get the equation:
$$ 64a - 2\cdot 192 + 240 = 48 $$
whose solution is indeed $a = 3$.

As an afterthought: there is another solution, although it might be an overkill. Use that the term at $x^2$ in polynomial $p$ is $p''(0)/2$. Your polynomial is:
$$ p(x) = (1+2x+ax^2)(2-x)^6$$
so you can compute easily enough:
$$ p'(x) = (2+2ax)(2-x)^6 + 6(1+2x+ax^2)(2-x)^5 $$
and then:
$$ p''(x) = 2a(2-x)^6 + 2 \cdot 6(2+2ax)(2-x)^5 + 30(1+2x+ax^2)(2-x)^4 $$
You can now plug in $x=0$:
$$ p''(0) = 2a \cdot 2^6 + 2 \cdot 6 \cdot 2 \cdot 2^5 + 30 \cdot 2^4 $$
On the other hand, you have
$$p''(0) = 2 \cdot 48$$
These two formulas for $p''(0)$ let you write down an equation for $a$.
A: 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 $$(1)(240)-(2)(192)+64a$$
A: 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. $\ \boxed{a=3}$
A: You need not expand completely as only low powers play a role:
$(2-x)^6 = 2^6-6\cdot 2^5\cdot x + \frac{6\cdot5}2\cdot 2^4\cdot x^2+\ldots = 64-192x+240x^2+\ldots$
where the dots represent anything involving $x^3$ or even higher powers.
After this
$(1+2x+ax^2)(2-x)^6 = (64-192x+240x^2+\ldots) + (128x-384x^2+\ldots)+ (64 a x^2+\ldots) = 64 -64 x + (-144+64a)x^2+\ldots$
