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Sagot :
To determine all potential rational roots of the polynomial [tex]\( f(x) = 15x^{11} - 6x^8 + x^3 - 4x + 3 \)[/tex] using the Rational Root Theorem, we proceed with the following steps:
1. Identify the constant term and the leading coefficient:
- The constant term (the term without [tex]\( x \)[/tex]) is [tex]\( 3 \)[/tex].
- The leading coefficient (the coefficient of the highest degree term) is [tex]\( 15 \)[/tex].
2. Find all factors of the constant term ([tex]\(3\)[/tex]):
- The factors of [tex]\( 3 \)[/tex] are [tex]\( \pm 1 \)[/tex] and [tex]\( \pm 3 \)[/tex].
3. Find all factors of the leading coefficient ([tex]\(15\)[/tex]):
- The factors of [tex]\( 15 \)[/tex] are [tex]\( \pm 1 \)[/tex], [tex]\( \pm 3 \)[/tex], [tex]\( \pm 5 \)[/tex], and [tex]\( \pm 15 \)[/tex].
4. Form all possible ratios of the factors of the constant term to the factors of the leading coefficient:
- The potential rational roots are given by the formula: [tex]\( \frac{\text{factor of constant term}}{\text{factor of leading coefficient}} \)[/tex].
5. List all unique ratios:
- Taking all combinations, we get the following ratios:
[tex]\[ \pm \frac{1}{1}, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15} \][/tex]
[tex]\[ \pm \frac{3}{1}, \pm \frac{3}{3}, \pm \frac{3}{5}, \pm \frac{3}{15} \][/tex]
6. Simplify the ratios and list the unique potential roots:
- Simplifying the above expressions, we obtain:
[tex]\[ \pm 1, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15} \][/tex]
[tex]\[ \pm 3, \pm 1, \pm \frac{3}{5}, \pm \frac{1}{5} \][/tex]
- Combining and removing duplicates gives the unique set of potential rational roots:
[tex]\[ \pm 1, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm 3, \pm \frac{3}{5}, \pm 5, \pm 15 \][/tex]
Thus, the list of all potential rational roots is:
[tex]\[ \pm \frac{1}{15}, \pm \frac{1}{5}, \pm \frac{1}{3}, \pm \frac{3}{5}, \pm 1, \pm \frac{5}{3}, \pm 3, \pm 5, \pm 15 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\pm \frac{1}{15}, \pm \frac{1}{5}, \pm \frac{1}{3}, \pm \frac{3}{5}, \pm 1, \pm \frac{5}{3}, \pm 3, \pm 5, \pm 15} \][/tex]
1. Identify the constant term and the leading coefficient:
- The constant term (the term without [tex]\( x \)[/tex]) is [tex]\( 3 \)[/tex].
- The leading coefficient (the coefficient of the highest degree term) is [tex]\( 15 \)[/tex].
2. Find all factors of the constant term ([tex]\(3\)[/tex]):
- The factors of [tex]\( 3 \)[/tex] are [tex]\( \pm 1 \)[/tex] and [tex]\( \pm 3 \)[/tex].
3. Find all factors of the leading coefficient ([tex]\(15\)[/tex]):
- The factors of [tex]\( 15 \)[/tex] are [tex]\( \pm 1 \)[/tex], [tex]\( \pm 3 \)[/tex], [tex]\( \pm 5 \)[/tex], and [tex]\( \pm 15 \)[/tex].
4. Form all possible ratios of the factors of the constant term to the factors of the leading coefficient:
- The potential rational roots are given by the formula: [tex]\( \frac{\text{factor of constant term}}{\text{factor of leading coefficient}} \)[/tex].
5. List all unique ratios:
- Taking all combinations, we get the following ratios:
[tex]\[ \pm \frac{1}{1}, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15} \][/tex]
[tex]\[ \pm \frac{3}{1}, \pm \frac{3}{3}, \pm \frac{3}{5}, \pm \frac{3}{15} \][/tex]
6. Simplify the ratios and list the unique potential roots:
- Simplifying the above expressions, we obtain:
[tex]\[ \pm 1, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15} \][/tex]
[tex]\[ \pm 3, \pm 1, \pm \frac{3}{5}, \pm \frac{1}{5} \][/tex]
- Combining and removing duplicates gives the unique set of potential rational roots:
[tex]\[ \pm 1, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm 3, \pm \frac{3}{5}, \pm 5, \pm 15 \][/tex]
Thus, the list of all potential rational roots is:
[tex]\[ \pm \frac{1}{15}, \pm \frac{1}{5}, \pm \frac{1}{3}, \pm \frac{3}{5}, \pm 1, \pm \frac{5}{3}, \pm 3, \pm 5, \pm 15 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\pm \frac{1}{15}, \pm \frac{1}{5}, \pm \frac{1}{3}, \pm \frac{3}{5}, \pm 1, \pm \frac{5}{3}, \pm 3, \pm 5, \pm 15} \][/tex]
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