For the zeros:

Put \(\displaystyle{P}{\left({x}\right)}={0}\)

So,

\(\displaystyle{P}{\left({x}\right)}={0}\)

\(\displaystyle\Rightarrow{3}{x}^{{{3}}}+{5}{x}^{{{2}}}-{2}{x}-{4}={0}\)

\(\displaystyle\Rightarrow{\left({x}+{1}\right)}{\left({3}{x}^{{{2}}}+{2}{x}-{4}\right)}={0}\)

Using the Zero Factor Principle: If \(\displaystyle{a}{b}={0}\) then \(\displaystyle{a}={0}\) or \(\displaystyle{b}={0}\)

So,

\(\displaystyle{x}+{1}={0}\) or \(\displaystyle{3}{x}^{{{2}}}+{2}{x}-{4}={0}\)

Now,

For a quadratic equation of the form \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}={0}\) the solution are

\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)

Therefore,

\(\displaystyle{x}+{1}={0}\Rightarrow{x}=-{1}\)

And

\(\displaystyle{3}{x}^{{{2}}}+{2}{x}-{4}={0}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{2}\pm\sqrt{{{2}^{{{2}}}-{4.3}{\left(-{4}\right)}}}}}{{{2}\cdot{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{2}\pm{2}\sqrt{{{13}}}}}{{{2}\cdot{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{1}\pm\sqrt{{{13}}}}}{{{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{1}+\sqrt{{{13}}}}}{{{3}}}},{\frac{{-{1}-\sqrt{{{13}}}}}{{{3}}}}\)

Hence, the real zeros of the given polynomial are:

\(\displaystyle{x}=-{1},{\frac{{-{1}+\sqrt{{{13}}}}}{{{3}}}},{\frac{{-{1}-\sqrt{{{13}}}}}{{{3}}}}\)

Put \(\displaystyle{P}{\left({x}\right)}={0}\)

So,

\(\displaystyle{P}{\left({x}\right)}={0}\)

\(\displaystyle\Rightarrow{3}{x}^{{{3}}}+{5}{x}^{{{2}}}-{2}{x}-{4}={0}\)

\(\displaystyle\Rightarrow{\left({x}+{1}\right)}{\left({3}{x}^{{{2}}}+{2}{x}-{4}\right)}={0}\)

Using the Zero Factor Principle: If \(\displaystyle{a}{b}={0}\) then \(\displaystyle{a}={0}\) or \(\displaystyle{b}={0}\)

So,

\(\displaystyle{x}+{1}={0}\) or \(\displaystyle{3}{x}^{{{2}}}+{2}{x}-{4}={0}\)

Now,

For a quadratic equation of the form \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}={0}\) the solution are

\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)

Therefore,

\(\displaystyle{x}+{1}={0}\Rightarrow{x}=-{1}\)

And

\(\displaystyle{3}{x}^{{{2}}}+{2}{x}-{4}={0}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{2}\pm\sqrt{{{2}^{{{2}}}-{4.3}{\left(-{4}\right)}}}}}{{{2}\cdot{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{2}\pm{2}\sqrt{{{13}}}}}{{{2}\cdot{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{1}\pm\sqrt{{{13}}}}}{{{3}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{1}+\sqrt{{{13}}}}}{{{3}}}},{\frac{{-{1}-\sqrt{{{13}}}}}{{{3}}}}\)

Hence, the real zeros of the given polynomial are:

\(\displaystyle{x}=-{1},{\frac{{-{1}+\sqrt{{{13}}}}}{{{3}}}},{\frac{{-{1}-\sqrt{{{13}}}}}{{{3}}}}\)