To determine the rational roots of the polynomial [tex]\( f(x) = 20x^4 + x^3 + 8x^2 + x - 12 \)[/tex], we use the Rational Root Theorem. This theorem states that any rational root of the polynomial, when written in its simplest form [tex]\( \frac{p}{q} \)[/tex], must have [tex]\( p \)[/tex] as a factor of the constant term and [tex]\( q \)[/tex] as a factor of the leading coefficient.
1. Identify the constant term and the leading coefficient:
- The constant term is [tex]\(-12\)[/tex].
- The leading coefficient is [tex]\(20\)[/tex].
2. Find the factors of the constant term and the leading coefficient:
- Factors of [tex]\(-12\)[/tex] are [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\)[/tex].
- Factors of [tex]\(20\)[/tex] are [tex]\(\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20\)[/tex].
3. List all possible rational roots:
- By combining the factors of the constant term with the factors of the leading coefficient, the potential rational roots are:
[tex]\[
\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm \frac{1}{20}, \pm 2, \pm \frac{2}{2}, \pm \frac{2}{4}, \pm \frac{2}{5}, \pm \frac{2}{10}, \pm \frac{2}{20}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{5}, \pm \frac{3}{10}, \pm \frac{3}{20}, \pm 4, \pm \frac{4}{2}, \pm \frac{4}{4}, \pm \frac{4}{5}, \pm \frac{4}{10}, \pm \frac{4}{20}, \pm 6, \pm \frac{6}{2}, \pm \frac{6}{4}, \pm \frac{6}{5}, \pm \frac{6}{10}, \pm \frac{6}{20}, \pm 12, \pm \frac{12}{2}, \pm \frac{12}{4}, \pm \frac{12}{5}, \pm \frac{12}{10}, \pm \frac{12}{20}.
\][/tex]
4. Verify the rational roots from the simplified list:
After testing all candidates, the valid rational roots are [tex]\(\frac{3}{4}\)[/tex] and [tex]\(-\frac{4}{5}\)[/tex]. Hence, we list all these validated roots:
The correct answer is:
[tex]\[
-\frac{4}{5} \text{ and } \frac{3}{4}
\][/tex]
