How may one solve problems over expressions like $(2+px)^6$ without the binomial theorem? A friend of mine posed a problem on a mathematics discord server.

The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.

I immediately thought of employing the binomial theorem, as what was required, actually. But, I decided to do another method, which led me to the question of how we can solve such problems without using the binomial theorem.
How can we solve this question without the binomial theorem or newton's method?

My Attempt:
Lemma: $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc.$$
Note that $$(2+px)^6=(2+px)^2(2+px)^2(2+px)^2$$ $$=((px)^2+4px+4)((px)^2+4px+4)((px)^2+4px+4)$$ Therefore, in the lemma, substitute $x\mapsto (px)^2$ and $a,b,c\mapsto 4px+4$. It follows $$(2+px)^6=(px)^6+12(px+1)(px)^4+48(px+1)^2(px)^2+64(px+1)^3.$$ We can ignore the first two terms since they contain no strict $x^2$ coefficient. Thus, upon expanding $$48(px+1)^2(px)^2+64(px+1)^3$$ it follows the coefficient of $x^2$ is $240p^2$. $$240p^2=60\tag{$p>0$}$$ $\therefore p=\frac 12$.

This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...
Let 
$$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$
Differentiate twice:
$$\begin{align}
p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2)
\\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x)
\end{align}$$
Evaluate at $x=0$:
$$30p^2\cdot 2^4=120\Rightarrow p^2=\frac{1}{4}\Rightarrow p\underset{p>0}{=}+\sqrt{\frac14}=\frac12.$$
