<s>[INST] Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
 \hline 
    x^2 &   \\ 
 \hline
     & 3 \\ 
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
 \hline 
    x^2 &  1x \\ 
 \hline
     3x& 3 \\ 
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
 $$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
 [/INST] Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.  
Suppose we have a quadratic $p(x)$ that factorizes as:
$$
p(x) = (x + s)(x + t)
$$
where we don't know what $s$ and $t$ are yet.  If we multiply out the brackets, then we get:
$$
p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st
$$
Now suppose that we are given $p(x)$ in the form
$$
p(x) = x^2 + bx + c
$$
In order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that
$$s+t=b$$
and
$$st=c$$
In your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$.  It shouldn't take you very long to realize that $s=3$ and $t=1$ will do.  Then you can immediately factorize the polynomial as $(x+3)(x+1)$.  
Slogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.

Just for interest
This method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'.  You might be interested in how we solve quadratic equations that don't have nice solutions.  For example, suppose we were trying to factorize
$$
x^2+x-1
$$
We want to find two numbers that add together to give $1$ and multiply together to give $-1$.  It turns out that the right two numbers are
$$
\frac{1+\sqrt{5}}{2}
$$
and
$$
\frac{1-\sqrt{5}}{2}
$$
How did I work those out?  Well, I'll show you a method that was invented by the ancient Babylonians.  Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.  
If we square the first equation, we get
$$
b^2 = (s+t)^2 = s^2 + 2st + t^2
$$
Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get:
$$
b^2-4c = s^2 - 2st + t^2
$$
But $s^2 - 2st + t^2=(s-t)^2$, so we may write
$$
s-t = \sqrt{b^2-4c}
$$
(Here, we are free to assume that $s\ge t$, so this is the positive square root.)
Now we can recover $s$ and $t$:
$$
s = \frac{(s+t) + (s-t)}{2} = \frac{b+\sqrt{b^2-4c}}{2}
$$
$$
t = \frac{(s+t) - (s-t)}{2} = \frac{b-\sqrt{b^2-4c}}{2}
$$
</s>