Question
Use the Rational Zero Theorem to determine all possible rational zeroes of f left parenthesis x right parenthesis equals 3 x cubed minus 2 x squared plus 5 x plus 9.
(1 point)
Responses
0, plus or minus one third, plus or minus 1, plus or minus 3
0, Image with alt text: plus or minus one third , Image with alt text: plus or minus 1 , Image with alt text: plus or minus 3
plus or minus one third, plus or minus 1, plus or minus 3, plus or minus 9
Image with alt text: plus or minus one third , Image with alt text: plus or minus 1 , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9
plus or minus 1 third, plus or minus 7 thirds, plus or minus 3, plus or minus 9
Image with alt text: plus or minus 1 third , Image with alt text: plus or minus 7 thirds , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9
plus or minus 1 third, plus or minus 3, plus or minus 9, plus or minus 27
Image with alt text: plus or minus 1 third , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9 , Image with alt text: plus or minus 27
(1 point)
Responses
0, plus or minus one third, plus or minus 1, plus or minus 3
0, Image with alt text: plus or minus one third , Image with alt text: plus or minus 1 , Image with alt text: plus or minus 3
plus or minus one third, plus or minus 1, plus or minus 3, plus or minus 9
Image with alt text: plus or minus one third , Image with alt text: plus or minus 1 , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9
plus or minus 1 third, plus or minus 7 thirds, plus or minus 3, plus or minus 9
Image with alt text: plus or minus 1 third , Image with alt text: plus or minus 7 thirds , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9
plus or minus 1 third, plus or minus 3, plus or minus 9, plus or minus 27
Image with alt text: plus or minus 1 third , Image with alt text: plus or minus 3 , Image with alt text: plus or minus 9 , Image with alt text: plus or minus 27
Answers
Answered by
GPT-4o mini
To determine the possible rational zeros of the polynomial function \( f(x) = 3x^3 - 2x^2 + 5x + 9 \) using the Rational Root Theorem, we follow these steps:
1. Identify the leading coefficient and the constant term of the polynomial.
- The leading coefficient \( a_n = 3 \) (coefficient of \( x^3 \)).
- The constant term \( a_0 = 9 \).
2. Determine the factors of the constant term and the leading coefficient.
- Factors of \( 9 \): \( \pm 1, \pm 3, \pm 9 \).
- Factors of \( 3 \): \( \pm 1, \pm 3 \).
3. Form all possible rational zeros by taking the factors of the constant term divided by the factors of the leading coefficient:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } a_0}{\text{Factors of } a_n} = \frac{\pm 1, \pm 3, \pm 9}{\pm 1, \pm 3}.
\]
Calculating the combinations:
- From \( \pm 1 \): The resulting rational zeros are \( \pm 1 \).
- From \( \pm 3 \): The resulting rational zeros are \( \pm 3 \).
- From \( \pm 9 \): The resulting rational zeros are \( \pm 9 \) (when divided by the factors of \( 3 \), we get \( \pm 9, \pm 3, \pm 1\) directly).
Additionally, dividing \( 9 \) by \( 3 \) gives \( \pm 3 \) and \( \pm 1 \), and dividing \( 1 \) by \( 3 \) gives \( \pm \frac{1}{3} \).
Thus, the complete list of possible rational zeros is:
\[
\pm 1, \pm 3, \pm 9, \pm \frac{1}{3}.
\]
So, the potential rational zeros are:
\[
\pm \frac{1}{3}, \pm 1, \pm 3, \pm 9.
\]
The correct response from the options provided is:
\[
\pm \frac{1}{3}, \pm 1, \pm 3, \pm 9.
\]
1. Identify the leading coefficient and the constant term of the polynomial.
- The leading coefficient \( a_n = 3 \) (coefficient of \( x^3 \)).
- The constant term \( a_0 = 9 \).
2. Determine the factors of the constant term and the leading coefficient.
- Factors of \( 9 \): \( \pm 1, \pm 3, \pm 9 \).
- Factors of \( 3 \): \( \pm 1, \pm 3 \).
3. Form all possible rational zeros by taking the factors of the constant term divided by the factors of the leading coefficient:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } a_0}{\text{Factors of } a_n} = \frac{\pm 1, \pm 3, \pm 9}{\pm 1, \pm 3}.
\]
Calculating the combinations:
- From \( \pm 1 \): The resulting rational zeros are \( \pm 1 \).
- From \( \pm 3 \): The resulting rational zeros are \( \pm 3 \).
- From \( \pm 9 \): The resulting rational zeros are \( \pm 9 \) (when divided by the factors of \( 3 \), we get \( \pm 9, \pm 3, \pm 1\) directly).
Additionally, dividing \( 9 \) by \( 3 \) gives \( \pm 3 \) and \( \pm 1 \), and dividing \( 1 \) by \( 3 \) gives \( \pm \frac{1}{3} \).
Thus, the complete list of possible rational zeros is:
\[
\pm 1, \pm 3, \pm 9, \pm \frac{1}{3}.
\]
So, the potential rational zeros are:
\[
\pm \frac{1}{3}, \pm 1, \pm 3, \pm 9.
\]
The correct response from the options provided is:
\[
\pm \frac{1}{3}, \pm 1, \pm 3, \pm 9.
\]