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What is coefficient of $x^2$ in polynomial $(1-4x)^6(1+3x)^8$ ? I know how to do it for $(1-4x)^6$ for example, but how to do it for the product of two polynomials?

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    Margarita: Do you know about how to accept an answer? Just checking. You can accept exactly one answer as the answer most helpful to you, and for each accept (and each time you accept an answer to any question you ask), you earn 2 pts. (You can accept one answer for each question, by clicking on the grey $\checkmark$ next to the answer you'd like to accept.) You can also upvote every helpful answer you obtain. – 2017-02-14

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Strategy:

  1. Compute $\alpha, \beta$ in $$(1-4x)^6=1+\alpha x + \beta x^2 +O(x^3)$$

  2. Compute $\delta, \gamma$ in $$(1+3x)^8=1+\delta x + \gamma x^2 + O(x^3)$$

  3. The answer you seek is $\beta+\gamma+\alpha\delta$.

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    right, although I don't think the big O notation is well used here. – 2017-01-25
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    Could you please explain what do you mean? – 2017-01-25
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    I should find for two polynomials separately ? – 2017-01-25
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    i know how to do it only by multinomial theorem – 2017-01-25
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    Why do you use $O(x^3)$? I don't think it is appropriate here. – 2017-01-25
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    @JorgeFernándezHidalgo, it captures all of the higher-order terms, and is a completely standard use of this notation. – 2017-01-25
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    Why do you say it captures the higher order terms? – 2017-01-25
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    Big O notation is used for $x\to\infty$ (which seems to be what you're thinking of), and also for $x\to 0$. – 2017-01-25
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    Oh ok, thanks for explaining yo me. – 2017-01-25
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If you have two polynomials $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ and $b_mx^m\dots b_0$ then the coefficient of $x^2$ in the product is just $b_2a_0+b_1a_1+b_0a_2$. You say that you can find each of these terms.

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    in the first one $(1-4x)^6 = ( 6 \choose n_1 n_2) 1 ^{n_1} (-4x)^{n_2}$ – 2017-01-25
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    then $x^2 = -4x^{n_2} $ so $n_2 = 2$ and $6!/2! = 360$ so coefficient is 360? – 2017-01-25
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    you are missing some steps, can you find $a_0,a_1,b_0$ and $b_1$? – 2017-01-25
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    what is $a_0 a_1...$ ? – 2017-01-25
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    if you know how to do it, could you explain please? – 2017-01-25
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Basically, you just need to find all of the terms of degree 0, 1, or 2 in each of the two factors, and see what products among them have the desired degree. That's because any higher order terms won't be of interest in the final product.

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Let $f(x) = (1-4x)^6(1+3x)^8 = uv$, where $u = (1-4x)^6$ and $v=(1+3x)^8$ \begin{align*} f''(x) &= u'' v + 2 u'v' + v''\\ &= 6 \cdot 5 \cdot (-4)^2 (1-4x)^4 (1+3x)^8 + 2 \cdot 6 \cdot (-4) \cdot 8 \cdot 3 (1-4x)^5 (1+3x)^7 \\ & \qquad + 8 \cdot 7 \cdot (3)^2 (1-4x)^6 (1+3x)^6 \end{align*} Hence coefficient of $x^2$ is $\frac{1}{2!}f''(0) = (480 - 1152 + 504)/2 = -84 $

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    We didn't solve such exercises in this way :( If you know how to do it by multionomial theorem, please explain to me – 2017-01-25
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    We have \begin{align*} (1-4x)^6 &= \binom{6}{0} + \binom{6}{1}(-4x) + \binom{6}{2}(-4x)^2 + \cdots \\ &= 1 - 24x + 240x^2 - \cdots\\ (1+3x)^8 &= \binom{8}{0} + \binom{8}{1}(3x) + \binom{8}{2}(3x)^2 + \cdots \\ &= 1 + 24x + 252x^2 + \cdots \end{align*} Hence \begin{align*} (1-4x)^6(1+3x)^8 &= (1-24x+ 240 x^2 + \cdots )(1+24x + 252 x^2 + \cdots) \end{align*} The coefficient of $x^2$ is hence $$ 252 - 576 + 240 = -84 $$ – 2017-01-26