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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, $a=3$.


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.

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You can use the binomial theorem to expand $(2-x)^6$ – Mhenni Benghorbal Sep 2 '12 at 15:17
up vote 6 down vote accepted

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 $a$.

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thanks for your wonderful answer. Is there a formula other than mulitplying each term to find the coeffiecents of a factored polynomal? – yiyi Sep 2 '12 at 12:51
could you explain <<Use that the term at x2 in polynomial p is p′′(0)/2.>> you mean the coefficent of $x^2$ of $p(x)$ is equal to the first derivative of $p(0)/2$? $p(x)=6x^2$ $p^{\prime}(0)/2 = 12(0) = 0$ – yiyi Sep 2 '12 at 13:06
misread it. Yes the 2nd derivative would be that. – yiyi Sep 2 '12 at 13:55

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 $$(1)(240)-(2)(192)+64a$$

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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. $\ \boxed{a=3}$

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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 $(1+2x+ax^2)(2-x)^6 = (64-192x+240x^2+\ldots) + (128x-384x^2+\ldots)+ (64 a x^2+\ldots) = 64 -64 x + (-144+64a)x^2+\ldots$

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