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...

Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...

Now factorise the rows and columns:
  $$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=1(x+3)
 $$


The last line ought to be $3x +3=3(x+1)$.
