Answer :
To solve the problem using the Rational Zeros Theorem, we need to determine the possible rational zeros of the polynomial [tex]\( P(x) = 12x^5 + 8x^3 - 5x - 8 \)[/tex].
The Rational Zeros Theorem states that the possible rational zeros of a polynomial are given by the ratios of the factors of the constant term to the factors of the leading coefficient. Let's identify these factors:
1. Constant Term: The constant term of the polynomial [tex]\( P(x) \)[/tex] is [tex]\(-8\)[/tex]. The factors of [tex]\(-8\)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 4, \pm 8 \][/tex]
2. Leading Coefficient: The leading coefficient of the polynomial [tex]\( P(x) \)[/tex] is [tex]\(12\)[/tex]. The factors of [tex]\(12\)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \][/tex]
3. Possible Rational Zeros: The possible rational zeros are the ratios of the factors of the constant term to the factors of the leading coefficient. We need to form all possible combinations of these ratios:
[tex]\[ \text{Possible rational zeros} = \frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}} \][/tex]
This yields the following combinations:
[tex]\[ \frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 1}{\pm 3}, \frac{\pm 1}{\pm 4}, \frac{\pm 1}{\pm 6}, \frac{\pm 1}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 2}{\pm 1}, \frac{\pm 2}{\pm 2}, \frac{\pm 2}{\pm 3}, \frac{\pm 2}{\pm 4}, \frac{\pm 2}{\pm 6}, \frac{\pm 2}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 4}{\pm 1}, \frac{\pm 4}{\pm 2}, \frac{\pm 4}{\pm 3}, \frac{\pm 4}{\pm 4}, \frac{\pm 4}{\pm 6}, \frac{\pm 4}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 8}{\pm 1}, \frac{\pm 8}{\pm 2}, \frac{\pm 8}{\pm 3}, \frac{\pm 8}{\pm 4}, \frac{\pm 8}{\pm 6}, \frac{\pm 8}{\pm 12} \][/tex]
Simplifying these combinations, we get the following unique possible rational zeros:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \][/tex]
[tex]\[ \pm 2, \pm \frac{2}{3}, \pm \frac{2}{4} \quad (= \pm \frac{1}{2}), \quad \pm \frac{2}{6} \quad (= \pm \frac{1}{3}), \quad \pm \frac{2}{12} \quad (= \pm \frac{1}{6}), \][/tex]
[tex]\[ \pm 4, \pm \frac{4}{2} \quad (= \pm 2), \quad \pm \frac{4}{3}, \quad \pm \frac{4}{4} \quad (= \pm 1), \quad \pm \frac{4}{6} \quad (= \pm \frac{2}{3}), \quad \pm \frac{4}{12} \quad (= \pm \frac{1}{3}), \][/tex]
[tex]\[ \pm 8, \pm \frac{8}{2} \quad (= \pm 4), \quad \pm \frac{8}{3}, \quad \pm \frac{8}{4} \quad (= \pm 2), \quad \pm \frac{8}{6} \quad (= \pm \frac{4}{3}), \quad \pm \frac{8}{12} \quad (= \pm \frac{2}{3}) \][/tex]
Collecting all unique values, we have:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 8, \pm \frac{8}{3} \][/tex]
Thus, the list of possible rational zeros for the polynomial [tex]\( P(x) = 12x^5 + 8x^3 - 5x - 8 \)[/tex] is:
[tex]\[ x = \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 8, \pm \frac{8}{3} \][/tex]
The Rational Zeros Theorem states that the possible rational zeros of a polynomial are given by the ratios of the factors of the constant term to the factors of the leading coefficient. Let's identify these factors:
1. Constant Term: The constant term of the polynomial [tex]\( P(x) \)[/tex] is [tex]\(-8\)[/tex]. The factors of [tex]\(-8\)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 4, \pm 8 \][/tex]
2. Leading Coefficient: The leading coefficient of the polynomial [tex]\( P(x) \)[/tex] is [tex]\(12\)[/tex]. The factors of [tex]\(12\)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \][/tex]
3. Possible Rational Zeros: The possible rational zeros are the ratios of the factors of the constant term to the factors of the leading coefficient. We need to form all possible combinations of these ratios:
[tex]\[ \text{Possible rational zeros} = \frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}} \][/tex]
This yields the following combinations:
[tex]\[ \frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 1}{\pm 3}, \frac{\pm 1}{\pm 4}, \frac{\pm 1}{\pm 6}, \frac{\pm 1}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 2}{\pm 1}, \frac{\pm 2}{\pm 2}, \frac{\pm 2}{\pm 3}, \frac{\pm 2}{\pm 4}, \frac{\pm 2}{\pm 6}, \frac{\pm 2}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 4}{\pm 1}, \frac{\pm 4}{\pm 2}, \frac{\pm 4}{\pm 3}, \frac{\pm 4}{\pm 4}, \frac{\pm 4}{\pm 6}, \frac{\pm 4}{\pm 12}, \][/tex]
[tex]\[ \frac{\pm 8}{\pm 1}, \frac{\pm 8}{\pm 2}, \frac{\pm 8}{\pm 3}, \frac{\pm 8}{\pm 4}, \frac{\pm 8}{\pm 6}, \frac{\pm 8}{\pm 12} \][/tex]
Simplifying these combinations, we get the following unique possible rational zeros:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \][/tex]
[tex]\[ \pm 2, \pm \frac{2}{3}, \pm \frac{2}{4} \quad (= \pm \frac{1}{2}), \quad \pm \frac{2}{6} \quad (= \pm \frac{1}{3}), \quad \pm \frac{2}{12} \quad (= \pm \frac{1}{6}), \][/tex]
[tex]\[ \pm 4, \pm \frac{4}{2} \quad (= \pm 2), \quad \pm \frac{4}{3}, \quad \pm \frac{4}{4} \quad (= \pm 1), \quad \pm \frac{4}{6} \quad (= \pm \frac{2}{3}), \quad \pm \frac{4}{12} \quad (= \pm \frac{1}{3}), \][/tex]
[tex]\[ \pm 8, \pm \frac{8}{2} \quad (= \pm 4), \quad \pm \frac{8}{3}, \quad \pm \frac{8}{4} \quad (= \pm 2), \quad \pm \frac{8}{6} \quad (= \pm \frac{4}{3}), \quad \pm \frac{8}{12} \quad (= \pm \frac{2}{3}) \][/tex]
Collecting all unique values, we have:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 8, \pm \frac{8}{3} \][/tex]
Thus, the list of possible rational zeros for the polynomial [tex]\( P(x) = 12x^5 + 8x^3 - 5x - 8 \)[/tex] is:
[tex]\[ x = \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 8, \pm \frac{8}{3} \][/tex]
