\((3x+1)(x+4)\)

\[3x^2+13x+4\]\[4\times 3 =12\]
We have to find the numbers whose sum is \(13\) and the product is \(12\). The numbers are \(12\) and \(1\).
$$
\begin{aligned}
&=3x^2+12x+x+4\\
&=3x(x+4)+1(x+4)\\
&=(3x+1)(x+4)
\end{aligned}
$$