What is the coefficient of $x^2$ in the expansion of $ (x+2)^4 \cdot (x+3)^5$? Can someone help me to solve this?
If I try to apply the binomial theorem, I don't know how to handle the two different exponentials.
Is there a way to simplify this to receive a pattern like $(a+b)^n$ that would give me an easy value for $n$?
 A: For the $x^2$ term of the whole product we need to take into account $3$ contributions, that is


*

*second degree factor from $(+2)^4$ and degree zero factor from $(+3)^5$

*zero degree factor from $(+2)^4$ and second degree factor from $(+3)^5$

*first degree factor from $(+2)^4$ and first degree factor from $(+3)^5$
A: Let $$f(x)=(x+2)^4(x+3)^5$$
$$=g(x)h(x)=a_0+a_1x+a_2x^2+...$$
then $2a_2$ is the constant term in the second derivative of $f$.
Using Leibnitz formula
$$f''(x)=g''(x)h(x)+2g'(x)h'(x)+g(x)h''(x)$$
$$=12(x+2)^2(x+3)^5+40(x+2)^3(x+3)^4+20(x+2)^4(x+3)^3$$
thus
$$2a_2=f''(0)=12.2^2.3^5+40.2^3.3^4+20.2^4.3^3$$
$$=2^2.3^3(108+240+80)$$
the desired coefficient is then
$$a_2=428\times 54=23112$$
A: After you expand the powers of the two monomials in the product, the only terms in those expansions that can contribute to the coefficient of $x^2$ in the full expansion have degree at most $2$. So, one way to attack this problem is to expand using the binomial theorem, discarding any cubic or higher terms: $$\left(\binom40 2^4+\binom41 2^3x+\binom42 2^2x^2+\cdots\right)\left(\binom50 3^5+\binom51 3^4x+\binom52 3^3x^2+\cdots\right).$$ Now pair up the surviving terms  so that each product ends up with a factor of $x^2$ and add up the coefficients of these pairwise products.
