To solve the polynomial \(\displaystyle{3}{x}^{{2}}{\left({5}{x}\right)}{\left(-{\frac{{{2}}}{{{3}}}}{x}^{{3}}\right)}\)

Apply the rule: \(\displaystyle{a}{\left(-{b}\right)}=-{a}{b}\)

\(\displaystyle{3}{\left({x}^{{2}}\right)}{\left({5}{x}\right)}{\left(-{\frac{{{2}}}{{{3}}}}{x}^{{3}}=-{\left({3}{x}^{{2}}\right)}{\left({5}{x}\right)}{\left({\frac{{{2}}}{{{3}}}}{x}^{{3}}\right)}\ {\left({a}{\left(-{b}\right)}=-{a}{b}\right)}\right.}\)

\(\displaystyle=-{3}{x}^{{{2}+{1}+{3}}}{\left({5}\right)}{\left({\frac{{{2}}}{{{3}}}}\right)}\ {\left({a}^{{b}}.{a}^{{c}}={a}^{{{b}+{c}}}\right)}\)

\(\displaystyle=-{3}{x}^{{6}}\cdot{5}\cdot{\frac{{{2}}}{{{3}}}}\)

\(\displaystyle=-{10}{x}^{{6}}\)

Therefore, \(\displaystyle{3}{\left({x}^{{2}}\right)}{\left({5}{x}\right)}{\left(-{\frac{{{2}}}{{{3}}}}{x}^{{3}}\right)}=-{10}{x}^{{6}}\)

Apply the rule: \(\displaystyle{a}{\left(-{b}\right)}=-{a}{b}\)

\(\displaystyle{3}{\left({x}^{{2}}\right)}{\left({5}{x}\right)}{\left(-{\frac{{{2}}}{{{3}}}}{x}^{{3}}=-{\left({3}{x}^{{2}}\right)}{\left({5}{x}\right)}{\left({\frac{{{2}}}{{{3}}}}{x}^{{3}}\right)}\ {\left({a}{\left(-{b}\right)}=-{a}{b}\right)}\right.}\)

\(\displaystyle=-{3}{x}^{{{2}+{1}+{3}}}{\left({5}\right)}{\left({\frac{{{2}}}{{{3}}}}\right)}\ {\left({a}^{{b}}.{a}^{{c}}={a}^{{{b}+{c}}}\right)}\)

\(\displaystyle=-{3}{x}^{{6}}\cdot{5}\cdot{\frac{{{2}}}{{{3}}}}\)

\(\displaystyle=-{10}{x}^{{6}}\)

Therefore, \(\displaystyle{3}{\left({x}^{{2}}\right)}{\left({5}{x}\right)}{\left(-{\frac{{{2}}}{{{3}}}}{x}^{{3}}\right)}=-{10}{x}^{{6}}\)