# Find all the real zeros of the polynomial. P(x)=3x^{3}+5x^{2}-2x-4

Find all the real zeros of the polynomial.
$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{3}}}+{5}{x}^{{{2}}}-{2}{x}-{4}$$

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i1ziZ
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:
$$\displaystyle{x}=-{1},{\frac{{-{1}+\sqrt{{{13}}}}}{{{3}}}},{\frac{{-{1}-\sqrt{{{13}}}}}{{{3}}}}$$