Solution
- The solution steps are given below:
[tex]\begin{gathered} 6x^4+2x^3-4x^2+2 \\ p\text{ is the set of all the factors of the constant term.} \\ p=\pm1,\pm2 \\ \\ q\text{ is the set of all the factors of the leading term} \\ q=\pm1,\pm2,\pm3,\pm6 \\ \\ \text{ Thus, the potential zeros are:} \\ \frac{p}{q}=\frac{1}{1},-\frac{1}{1},\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{1}{6},-\frac{1}{6},\frac{2}{1},-\frac{2}{1},\frac{2}{3},-\frac{2}{3} \\ \\ \text{ Thus, this set of potential zeros is given as:} \\ \frac{p}{q}=\lbrace\pm\frac{1}{6},\pm\frac{1}{3},\pm\frac{1}{2},\pm\frac{2}{3},\pm1,\pm2\rbrace \end{gathered}[/tex]
Final Answer
The answer is OPTION C
