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...