Factor each expression completely:

(a) $2( x+3) ( x-2) ^{3}+( x+3) ^{2}( 3) ( x-2) ^{2}$

(b) $\dfrac{4}{3}x^{1/3}( 2x+1) +2x^{4/3}$

Solution (a) In expression (a), $( x+3)$ and $( x-2) ^{2}$ are common factors, factors found in each term. Factor them out. $$\begin{array}{rcl@{\qquad\hspace*{-3pt}}l} &&2 ( x+3) ( x-2) ^{3}+ ( x+3) ^{2} ( 3) ( x-2) ^{2}\\[5pt] &&\quad = ( x+3) (x-2) ^{2} [ 2 ( x-2) +3 ( x+3) ] & {\color{#0066A7}{\hbox{Factor out \(\hbox{(}x+3\hbox{)}\,\hbox{(}x-2\hbox{)}^2\).}}} \\[5pt] &&\quad = ( x+3) ( x-2) ^{2} ( 5x+5) & {\color{#0066A7}{\hbox{Simplify.}}}\\[5pt] &&\quad = 5 ( x+3) ( x-2) ^{2} (x+1) & {\color{#0066A7}{\hbox{Factor out 5.}}} \end{array}$$

(b) We begin by writing the term $2x^{4/3}$ as a fraction with a denominator of $3$. $$\begin{array}{rcl@{\qquad\hspace*{-5.5pt}}l} \dfrac{4}{3}x^{1/3} ( 2x+1) +2x^{4/3} &=&\dfrac{4x^{1/3} (2x+1) }{3}+\dfrac{6x^{4/3}}{3}\\[9pt] &=&\dfrac{4x^{1/3} ( 2x+1) +6x^{4/3}}{3} &{\color{#0066A7}{\hbox{Add the two fractions.}}}\\[9pt] &=&\dfrac{2x^{1/3} [ 2 ( 2x+1) +3x] }{3} &{\color{#0066A7}{\hbox{2 and \(x^{1/3}\) are common factors.}}}\\[9pt] &=&\dfrac{2x^{1/3} (7x+2) }{3} & {\color{#0066A7}{\hbox{Simplify.}}} \end{array}$$