How can I find the coefficient of x when the power is greater than the powers of 2 brackets using binomial expansion? I have been given this question:
Find the coefficient of $x^{13}$ in the expansion of $(1 + 2x)^4(2 + x)^{10}$.
I know how I would find $x^4$ or lower degrees, but I am unsure how to approach this, as neither term has a $x^{13}$, and x is a prime number so it can't just be 2 terms multiplied (as neither bracket has a power of 13).
Where do I start with this?
 This is revision rather than homework, but hints would be appreciated.
 A: $(1+2x)^4$ give out terms containing $x$ with power $0,1,2,3,4$ similarly $(2+x)^{10}$ give out term containing $x$ with power $0$ to $10$ now pick up exponents from first and second such that these sum up to 13 along with coefficient.           ATP,
$$(10{C_0}x^{10})(4C_3(2x)^3)+2(10C_1x^9)(4C_4(2x)^4)$$ and these will give out coefficient.
A: If you expand the two terms, the first one will give you terms from $1$ to $16x^4$.  The second will give you terms from $2^{10}$ to $x^{10}$.  When you multiply them, the only two ways to get $x^{13}$ is to use the $x^3$ term from the first and the $x^{10}$ term from the second or to use the $x^4$ from the first and the $x^9$ from the second.  Evaluate the coefficients of each of these terms, multiply, and add.
A: Since $x^{10}\cdot x^3=x^{13}$ and $x^{9}\cdot x^4=x^{13}$, we just have to find the coefficients of $x^4,x^3$ in $(1+2x)^4$ and the coefficients of $x^{10},x^9$ in $(2+x)^{10}$. To do this you can use binomial thereom:
$$x^3:{4 \choose 3}(2x)^3=32x^3$$
$$x^4:{4 \choose 0}(2x)^4=16x^4$$
$$x^{10}:{10 \choose 0}(x)^{10}=x^{10}$$
$$x^9:{10 \choose 1}2(x)^9=20x^9.$$
So we have $20x^9\cdot 16x^4+32x^3\cdot x^{10}=352x^{13}. $
A: There is a useful notation, the so-called coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a polynomial or a series. With this notation you can do some kind of bookkeeping of the coefficients you need from the binomials.

We obtain
  \begin{align*}
[x^{13}]&(1+2x)^4(2+x)^{10}\\
&=[x^{13}]\left(\sum_{j=0}^4\binom{4}{j}2^jx^j\right)(2+x)^{10}\tag{1}\\
&=\left(\sum_{j=0}^4\binom{4}{j}2^j[x^{13-j}]\right)(2+x)^{10}\tag{2}\\
&=\left(\binom{4}{3}2^3[x^{10}]+\binom{4}{4}2^4[x^9]\right)(2+x)^{10}\tag{3}\\
&=\left(32[x^{10}]+16[x^9]\right)\sum_{j=0}^{10}\binom{10}{j}2^{10-j}x^j\tag{4}\\
&=32\binom{10}{10}2^0+16\binom{10}{9}2^1\tag{5}\\
&=352
\end{align*}

Comment: 


*

*In (1) we expand the left binomial

*In (2) we use the linearity of the coefficient of Operator and $[x^n]x^kp(x)=[x^{n-k}]p(x)$

*In (3) we see that only $j=3$ and $j=4$ provide a contribution

*In (4) we expand the other binomial expression

*In (5) we take the coefficients of $x^{10}$ and $x^9$
