# Find all rational zeros of the polynomial, and write the polynomial in factored form. P(x)=3x^{5}-14x^{4}-14x^{3}+36x^{2}+43x+10

Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{14}{x}^{{{4}}}-{14}{x}^{{{3}}}+{36}{x}^{{{2}}}+{43}{x}+{10}$$

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Step 1
Given polynomial is, $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{14}{x}^{{{4}}}-{14}{x}^{{{3}}}+{36}{x}^{{{2}}}+{43}{x}+{10}$$
By using rational root theorem,
$$-\frac{1}{1} \text{is the root of the expression, so factor out}\ x+1$$.
$$\displaystyle{\frac{{{3}{x}^{{{5}}}-{14}{x}^{{{4}}}-{14}{x}^{{{3}}}+{36}{x}^{{{2}}}+{43}{x}+{10}}}{{{x}+{1}}}}={3}{x}^{{{4}}}-{17}{x}^{{{3}}}+{3}{x}^{{{2}}}+{33}{x}+{10}$$
$$=(x+1)(x+1)(x-2)(3x+1)(x-5)$$
$$\displaystyle={\left({x}+{1}\right)}^{{{2}}}{\left({x}-{2}\right)}{\left({3}{x}+{1}\right)}{\left({x}-{5}\right)}$$
This is the factored form of the polynomial.
Step 2
and, the zeros are,
$$\displaystyle{3}{x}^{{{5}}}-{14}{x}^{{{4}}}-{14}{x}^{{{3}}}+{36}{x}^{{{2}}}+{43}{x}+{10}={0}$$
$$\displaystyle{\left({x}+{1}\right)}^{{{2}}}{\left({x}-{2}\right)}{\left({3}{x}+{1}\right)}{\left({x}-{5}\right)}={0}$$
$$x+1=0\ or\ x-2=0\ or\ 3x+1=0\ or\ x-5=0$$
$$\displaystyle{x}=-{1},{x}={2},{x}=-{\frac{{{1}}}{{{3}}}},{x}={5}$$