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Sagot :
To determine the possible rational zeros of the polynomial \( f(x) = 7x^4 + 3x^3 + 9x^2 + x - 3 \), we will use the Rational Zeros Theorem. This theorem states that any rational zero of a polynomial is of the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term (the term without \( x \)) and \( q \) is a factor of the leading coefficient (the coefficient of the highest degree term).
Let's identify the needed components in our polynomial \( f(x) = 7x^4 + 3x^3 + 9x^2 + x - 3 \):
1. Constant term: \(-3\)
2. Leading coefficient: \(7\)
### Step 1: Identify the factors of the constant term \(-3\)
The factors of \(-3\) are:
[tex]\[ \{-3, -1, 1, 3\} \][/tex]
### Step 2: Identify the factors of the leading coefficient \(7\)
The factors of \(7\) are:
[tex]\[ \{-7, -1, 1, 7\} \][/tex]
### Step 3: Form all possible rational zeros \(\frac{p}{q}\)
Now we create all possible ratios of the factors of the constant term to the factors of the leading coefficient. These ratios include all combinations of \( \frac{p}{q} \) and must be simplified and listed without repetition.
[tex]\[ \frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \][/tex]
Here are the possible ratios:
[tex]\[ \begin{align*} \frac{-3}{-7}, \frac{-3}{-1}, \frac{-3}{1}, \frac{-3}{7}, \\ \frac{-1}{-7}, \frac{-1}{-1}, \frac{-1}{1}, \frac{-1}{7}, \\ \frac{1}{-7}, \frac{1}{-1}, \frac{1}{1}, \frac{1}{7}, \\ \frac{3}{-7}, \frac{3}{-1}, \frac{3}{1}, \frac{3}{7} \end{align*} \][/tex]
Next, we simplify these fractions:
[tex]\[ \begin{align*} \frac{-3}{-7} &= \frac{3}{7}, &\frac{-3}{-1} &= 3, &\frac{-3}{1} &= -3, &\frac{-3}{7} &= -\frac{3}{7}, \\ \frac{-1}{-7} &= \frac{1}{7}, &\frac{-1}{-1} &= 1, &\frac{-1}{1} &= -1, &\frac{-1}{7} &= -\frac{1}{7}, \\ \frac{1}{-7} &= -\frac{1}{7}, &\frac{1}{-1} &= -1, &\frac{1}{1} &= 1, &\frac{1}{7} &= \frac{1}{7}, \\ \frac{3}{-7} &= -\frac{3}{7}, &\frac{3}{-1} &= -3, &\frac{3}{1} &= 3, &\frac{3}{7} &= \frac{3}{7} \end{align*} \][/tex]
### Step 4: List unique possible rational zeros
Removing duplicates from these results, we get the final list of potential rational zeros:
[tex]\[ \left\{-3, -1, -\frac{3}{7}, -\frac{1}{7}, \frac{1}{7}, \frac{3}{7}, 1, 3\right\} \][/tex]
Therefore, the list of all possible rational zeros of the polynomial \( f(x) = 7x^4 + 3x^3 + 9x^2 + x - 3 \) is:
[tex]\[ \left\{ \pm 1, \pm 3, \pm \frac{1}{7}, \pm \frac{3}{7} \right\} \][/tex]
Let's identify the needed components in our polynomial \( f(x) = 7x^4 + 3x^3 + 9x^2 + x - 3 \):
1. Constant term: \(-3\)
2. Leading coefficient: \(7\)
### Step 1: Identify the factors of the constant term \(-3\)
The factors of \(-3\) are:
[tex]\[ \{-3, -1, 1, 3\} \][/tex]
### Step 2: Identify the factors of the leading coefficient \(7\)
The factors of \(7\) are:
[tex]\[ \{-7, -1, 1, 7\} \][/tex]
### Step 3: Form all possible rational zeros \(\frac{p}{q}\)
Now we create all possible ratios of the factors of the constant term to the factors of the leading coefficient. These ratios include all combinations of \( \frac{p}{q} \) and must be simplified and listed without repetition.
[tex]\[ \frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \][/tex]
Here are the possible ratios:
[tex]\[ \begin{align*} \frac{-3}{-7}, \frac{-3}{-1}, \frac{-3}{1}, \frac{-3}{7}, \\ \frac{-1}{-7}, \frac{-1}{-1}, \frac{-1}{1}, \frac{-1}{7}, \\ \frac{1}{-7}, \frac{1}{-1}, \frac{1}{1}, \frac{1}{7}, \\ \frac{3}{-7}, \frac{3}{-1}, \frac{3}{1}, \frac{3}{7} \end{align*} \][/tex]
Next, we simplify these fractions:
[tex]\[ \begin{align*} \frac{-3}{-7} &= \frac{3}{7}, &\frac{-3}{-1} &= 3, &\frac{-3}{1} &= -3, &\frac{-3}{7} &= -\frac{3}{7}, \\ \frac{-1}{-7} &= \frac{1}{7}, &\frac{-1}{-1} &= 1, &\frac{-1}{1} &= -1, &\frac{-1}{7} &= -\frac{1}{7}, \\ \frac{1}{-7} &= -\frac{1}{7}, &\frac{1}{-1} &= -1, &\frac{1}{1} &= 1, &\frac{1}{7} &= \frac{1}{7}, \\ \frac{3}{-7} &= -\frac{3}{7}, &\frac{3}{-1} &= -3, &\frac{3}{1} &= 3, &\frac{3}{7} &= \frac{3}{7} \end{align*} \][/tex]
### Step 4: List unique possible rational zeros
Removing duplicates from these results, we get the final list of potential rational zeros:
[tex]\[ \left\{-3, -1, -\frac{3}{7}, -\frac{1}{7}, \frac{1}{7}, \frac{3}{7}, 1, 3\right\} \][/tex]
Therefore, the list of all possible rational zeros of the polynomial \( f(x) = 7x^4 + 3x^3 + 9x^2 + x - 3 \) is:
[tex]\[ \left\{ \pm 1, \pm 3, \pm \frac{1}{7}, \pm \frac{3}{7} \right\} \][/tex]
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