Not really a lot to do with this problem. All we need to do is use the formula from the Binomial Theorem to do the expansion. Here is that work.

\[\begin{align*}{\left( {4 + 3x} \right)^5} & = \sum\limits_{i = 0}^5 { {5 \choose i} {4^{5 - i}}{{\left( {3x} \right)}^i}} \\ & = {5 \choose 0}\left( {{4^5}} \right) + {5 \choose 1}\left( {{4^4}} \right){\left( {3x} \right)^1} + {5 \choose 2} \left( {{4^3}} \right){\left( {3x} \right)^2} + {5 \choose 3}\left( {{4^2}} \right){\left( {3x} \right)^3} + {5 \choose 4}\left( {{4^1}} \right){\left( {3x} \right)^4}\\ & \hspace{1.5in} + {5 \choose 5} \left(3x\right)^{5} \\ & = {4^5} + \left( 5 \right)\left( {{4^4}} \right)\left( {3x} \right) + \frac{{5\left( 4 \right)}}{{2!}}\left( {{4^3}} \right){\left( {3x} \right)^2} + \frac{{5\left( 4 \right)\left( 3 \right)}}{{3!}}\left( {{4^2}} \right){\left( {3x} \right)^3} + \left( 5 \right)\left( 4 \right){\left( {3x} \right)^4} + {\left( {3x} \right)^5}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{1024 + 3840x + 5760{x^2} + 4320{x^3} + 1620{x^4} + 243{x^5}}}\end{align*}\]