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

Khadija Wells 2021-03-20 Answered
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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Expert Answer

davonliefI
Answered 2021-03-22 Author has 14157 answers
Given
\(\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}\)
Answer
\(\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{7}{x}^{{{3}}}-{12}{x}^{{{2}}}+{3}{x}+{2}\)
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),
\(\displaystyle{p}{\left({x}\right)}={\left({x}+{1}\right)}{\left({x}+{\frac{{{1}}}{{{3}}}}\right)}{\left({x}-{\frac{{{1}}}{{{2}}}}\right)}{\left({x}-{2}\right)}\)
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