Factorise $f(x) = x^3+4x^2 + 3x$

Had never heard of the box method before I saw you use it!

When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.

Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.

This works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\lambda_1, \cdots, \lambda_n$, then the polynomial is equal to $(x - \lambda_1)\cdots(x - \lambda_n)$

Of course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.

First obvious thing is that $\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.

Now comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.

I wouldn't know if this really qualifies as an answer, but what you're asking for is essentially an opinion of methodology.

First of all, your factorisation is correct: you can check the result simply by doing the multiplication.

That said, it looks like you could have saved a lot of effort by using a couple of different techniques.

To begin with, the polynomial $x^3+4x^2+3x$ is clearly divisible by $x$ (as you noted), so we factor it out to get $x(x^2+4x+3)$.

Here is where it gets interesting: to factor the degree 2 polynomial, I usually take one of these two roads.

First: observe that $(x+a)(x+b)=x^2+(a+b)x+ab$, so if we can guess two numbers $a$ and $b$ such that, in our case, we have $ab=3$ and $a+b=4$, we are done.

In this case you might have been able to spot that $3\cdot 1=3$ and $3+1=4$, so $(x+1)(x+3)$ is the factorization we're looking for. Trust me, it gets easier with practice and it's a huge time- and effort-saver for simple degree 2 polynomials with integer coefficients.

Else we could try and guess just one root, and then use polynomial long division. This has the advantage of working with any degree of polynomials, as long as you can guess one root. I usually try $1$, $-1$ and maybe $2$, but if those don't work I'm better off using some other technique.

And finally, for degree 2 polynomials you have the general formula to find the roots, which has the advantage of always working (in the sense that it will always find all real roots there are to find), but can get messy and is, in my honest opinion, rather boring.

Now, this is not to say that any of what you did is incorrect or even strictly less efficient: it is ultimately a matter of personal taste, but I feel that having as many tools as possible in your pocket can only do you good.