# Find all rational zeros of the polynomial, and write the polynomial in factored form. P(x)=6x^{4}-7x^{3}-12x^{2}+3x+2

Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}$$

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davonliefI
Given
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}$$
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}$$
use long division to find zeros of p(x)
from the long division, we get
x=-1, x=2 and $$\displaystyle{6}{x}^{{{2}}}-{x}-{1}={0}$$
$$\displaystyle{x}={\frac{{-{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{{2}}}-{4}{\left({6}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({6}\right)}}}}$$
$$\displaystyle={\frac{{{1}\pm\sqrt{{{1}+{24}}}}}{{{12}}}}$$
$$\displaystyle={\frac{{{1}\pm\sqrt{{{25}}}}}{{{12}}}}$$
$$\displaystyle={\frac{{{1}\pm{5}}}{{{12}}}}$$
$$\displaystyle{x}={\frac{{{6}}}{{{12}}}}\ {\quad\text{and}\quad}\ {x}={\frac{{-{4}}}{{{12}}}}$$
$$\displaystyle{x}={\frac{{{1}}}{{{2}}}}\ {\quad\text{and}\quad}\ {x}=-{\frac{{{1}}}{{{3}}}}$$
Therefore the zeros of p(x) is x =-1, $$\displaystyle-{\frac{{{1}}}{{{3}}}},{\frac{{{1}}}{{{2}}}}$$ and 2
Factor of p(x),
$$\displaystyle{p}{\left({x}\right)}={\left({x}+{1}\right)}{\left({x}+{\frac{{{1}}}{{{3}}}}\right)}{\left({x}-{\frac{{{1}}}{{{2}}}}\right)}{\left({x}-{2}\right)}$$