To determine the rational roots of the polynomial [tex]\( f(x) = 20x^4 + x^3 + 8x^2 + x - 12 \)[/tex], we will use the Rational Root Theorem. According to this theorem, any potential rational root of the polynomial [tex]\( f(x) \)[/tex] can be expressed as [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (in this case, -12), and [tex]\( q \)[/tex] is a factor of the leading coefficient (in this case, 20).
The factors of -12 are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \][/tex]
The factors of 20 are:
[tex]\[ \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \][/tex]
To find all possible rational roots, we consider all combinations of [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is any factor of -12 and [tex]\( q \)[/tex] is any factor of 20. This means we have a set of potential rational roots:
[tex]\[
\left\{ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm \frac{1}{20}, \pm 2, \pm \frac{2}{5}, \pm \frac{2}{10}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{5}, \pm 4, \pm \frac{4}{5}, \pm 6, \pm \frac{6}{5}, \pm 12, \pm \frac{12}{5}, \ldots \right\}
\][/tex]
Next, we evaluate each potential root by substituting it into the polynomial [tex]\( f(x) \)[/tex] to check if it equals zero.
When evaluated through the polynomial, we find that the rational roots of [tex]\( f(x) = 20x^4 + x^3 + 8x^2 + x - 12 \)[/tex] are:
[tex]\[
-\frac{4}{5} \quad \text{and} \quad \frac{3}{4}
\][/tex]
So, we can conclude that the rational roots are [tex]\( -\frac{4}{5} \)[/tex] and [tex]\( \frac{3}{4} \)[/tex].
Among the choices provided, the correct answer is:
[tex]\[
-\frac{4}{5} \text{ and } \frac{3}{4}
\][/tex]
Therefore, the correct choice is:
[tex]\[ \text{Choice D: } -\frac{4}{5} \text{ and } \frac{3}{4} \][/tex]
