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Sagot :
To find the set of possible rational zeros for the polynomial function [tex]\( f(x) = 5x^3 - 8x^2 + 3 \)[/tex], we can apply the Rational Root Theorem. This theorem states that any rational root, written in the 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 (the term without [tex]\(x\)[/tex]) in [tex]\( f(x) = 5x^3 - 8x^2 + 3 \)[/tex] is 3.
- The leading coefficient (the coefficient of the highest power of [tex]\(x\)[/tex]) is 5.
2. Find the factors of the constant term (3):
- The factors of 3 are [tex]\(\pm 1, \pm 3\)[/tex].
3. Find the factors of the leading coefficient (5):
- The factors of 5 are [tex]\(\pm 1, \pm 5\)[/tex].
4. Form all possible fractions [tex]\(\frac{p}{q}\)[/tex] where [tex]\(p\)[/tex] is a factor of the constant term and [tex]\(q\)[/tex] is a factor of the leading coefficient:
- List of [tex]\(\frac{p}{q}\)[/tex] combinations:
[tex]\[ \frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 5}, \frac{\pm 3}{\pm 1}, \frac{\pm 3}{\pm 5} \][/tex]
5. Simplify the fractions to obtain all unique values:
[tex]\[ \begin{align*} \frac{1}{1} &= 1 \\ \frac{-1}{1} &= -1 \\ \frac{1}{-1} &= 1 \\ \frac{-1}{-1} &= -1 \\ \frac{1}{5} &= 0.2 \\ \frac{-1}{5} &= -0.2 \\ \frac{1}{-5} &= -0.2 \\ \frac{-1}{-5} &= 0.2 \\ \frac{3}{1} &= 3 \\ \frac{-3}{1} &= -3 \\ \frac{3}{-1} &= -3 \\ \frac{-3}{-1} &= 3 \\ \frac{3}{5} &= 0.6 \\ \frac{-3}{5} &= -0.6 \\ \frac{3}{-5} &= -0.6 \\ \frac{-3}{-5} &= 0.6 \\ \end{align*} \][/tex]
Combining all unique values, we get the set of possible rational zeros:
[tex]\[ \{-3, -1, -\frac{3}{5}, -\frac{1}{5}, 0.2, 0.6, \frac{1}{5}, \frac{3}{5}, 3, 1\} \][/tex]
However, typically in simpler forms for rational expressions, we combine and remove duplicates, ensuring clarity:
[tex]\[ \{-3, -1, -0.6, -0.2, 0.2, 0.6, 1, 3 \} = \left\{-3, -1, -\frac{3}{5}, -\frac{1}{5}, \frac{1}{5}, \frac{3}{5}, 1, 3\right\} \][/tex]
Therefore, the set of possible rational zeros is:
[tex]\[ \{-3, -1, -\frac{3}{5}, -\frac{1}{5}, \frac{1}{5}, \frac{3}{5}, 1, 3\} \][/tex]
1. Identify the constant term and the leading coefficient:
- The constant term (the term without [tex]\(x\)[/tex]) in [tex]\( f(x) = 5x^3 - 8x^2 + 3 \)[/tex] is 3.
- The leading coefficient (the coefficient of the highest power of [tex]\(x\)[/tex]) is 5.
2. Find the factors of the constant term (3):
- The factors of 3 are [tex]\(\pm 1, \pm 3\)[/tex].
3. Find the factors of the leading coefficient (5):
- The factors of 5 are [tex]\(\pm 1, \pm 5\)[/tex].
4. Form all possible fractions [tex]\(\frac{p}{q}\)[/tex] where [tex]\(p\)[/tex] is a factor of the constant term and [tex]\(q\)[/tex] is a factor of the leading coefficient:
- List of [tex]\(\frac{p}{q}\)[/tex] combinations:
[tex]\[ \frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 5}, \frac{\pm 3}{\pm 1}, \frac{\pm 3}{\pm 5} \][/tex]
5. Simplify the fractions to obtain all unique values:
[tex]\[ \begin{align*} \frac{1}{1} &= 1 \\ \frac{-1}{1} &= -1 \\ \frac{1}{-1} &= 1 \\ \frac{-1}{-1} &= -1 \\ \frac{1}{5} &= 0.2 \\ \frac{-1}{5} &= -0.2 \\ \frac{1}{-5} &= -0.2 \\ \frac{-1}{-5} &= 0.2 \\ \frac{3}{1} &= 3 \\ \frac{-3}{1} &= -3 \\ \frac{3}{-1} &= -3 \\ \frac{-3}{-1} &= 3 \\ \frac{3}{5} &= 0.6 \\ \frac{-3}{5} &= -0.6 \\ \frac{3}{-5} &= -0.6 \\ \frac{-3}{-5} &= 0.6 \\ \end{align*} \][/tex]
Combining all unique values, we get the set of possible rational zeros:
[tex]\[ \{-3, -1, -\frac{3}{5}, -\frac{1}{5}, 0.2, 0.6, \frac{1}{5}, \frac{3}{5}, 3, 1\} \][/tex]
However, typically in simpler forms for rational expressions, we combine and remove duplicates, ensuring clarity:
[tex]\[ \{-3, -1, -0.6, -0.2, 0.2, 0.6, 1, 3 \} = \left\{-3, -1, -\frac{3}{5}, -\frac{1}{5}, \frac{1}{5}, \frac{3}{5}, 1, 3\right\} \][/tex]
Therefore, the set of possible rational zeros is:
[tex]\[ \{-3, -1, -\frac{3}{5}, -\frac{1}{5}, \frac{1}{5}, \frac{3}{5}, 1, 3\} \][/tex]
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