Question

list all of the possible rational zeros of each function. 23. $f(x) = 3x^4 - x^3 + 4$ 24. $f(x) = x^4 + 7x^3 - 15$ 25. $f(x) = x^3 - 6x^2 - 8x + 24$

Explanation:

To find the possible rational zeros of a polynomial function \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \), we use the Rational Root Theorem. This theorem states that any possible rational zero, \( \frac{p}{q} \), is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). In other words:

\[
\text{Possible rational zeros} = \frac{\text{Factors of } a_0}{\text{Factors of } a_n}
\]

Problem 23: \( f(x) = 3x^4 - x^3 + 4 \)

• Leading coefficient \( a_n = 3 \), so factors of \( 3 \) are \( \pm 1, \pm 3 \).
• Constant term \( a_0 = 4 \), so factors of \( 4 \) are \( \pm 1, \pm 2, \pm 4 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } 4}{\text{Factors of } 3} = \frac{\pm 1, \pm 2, \pm 4}{\pm 1, \pm 3}
\]

Simplify each fraction:
\[
\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}
\]

Problem 24: \( f(x) = x^4 + 7x^3 - 15 \)

• Leading coefficient \( a_n = 1 \), so factors of \( 1 \) are \( \pm 1 \).
• Constant term \( a_0 = -15 \), so factors of \( 15 \) are \( \pm 1, \pm 3, \pm 5, \pm 15 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } -15}{\text{Factors of } 1} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1}
\]

Simplify each fraction:
\[
\pm 1, \pm 3, \pm 5, \pm 15
\]

Problem 25: \( f(x) = x^3 - 6x^2 - 8x + 24 \)

• Leading coefficient \( a_n = 1 \), so factors of \( 1 \) are \( \pm 1 \).
• Constant term \( a_0 = 24 \), so factors of \( 24 \) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } 24}{\text{Factors of } 1} = \frac{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24}{\pm 1}
\]

Simplify each fraction:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24
\]

Final Answers:

1. Possible rational zeros: \( \boldsymbol{\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}} \)
2. Possible rational zeros: \( \boldsymbol{\pm 1, \pm 3, \pm 5, \pm 15} \)
3. Possible rational zeros: \( \boldsymbol{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24} \)

Answer:

To find the possible rational zeros of a polynomial function \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \), we use the Rational Root Theorem. This theorem states that any possible rational zero, \( \frac{p}{q} \), is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). In other words:

\[
\text{Possible rational zeros} = \frac{\text{Factors of } a_0}{\text{Factors of } a_n}
\]

Problem 23: \( f(x) = 3x^4 - x^3 + 4 \)

• Leading coefficient \( a_n = 3 \), so factors of \( 3 \) are \( \pm 1, \pm 3 \).
• Constant term \( a_0 = 4 \), so factors of \( 4 \) are \( \pm 1, \pm 2, \pm 4 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } 4}{\text{Factors of } 3} = \frac{\pm 1, \pm 2, \pm 4}{\pm 1, \pm 3}
\]

Simplify each fraction:
\[
\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}
\]

Problem 24: \( f(x) = x^4 + 7x^3 - 15 \)

• Leading coefficient \( a_n = 1 \), so factors of \( 1 \) are \( \pm 1 \).
• Constant term \( a_0 = -15 \), so factors of \( 15 \) are \( \pm 1, \pm 3, \pm 5, \pm 15 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } -15}{\text{Factors of } 1} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1}
\]

Simplify each fraction:
\[
\pm 1, \pm 3, \pm 5, \pm 15
\]

Problem 25: \( f(x) = x^3 - 6x^2 - 8x + 24 \)

• Leading coefficient \( a_n = 1 \), so factors of \( 1 \) are \( \pm 1 \).
• Constant term \( a_0 = 24 \), so factors of \( 24 \) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \).

Using the Rational Root Theorem:
\[
\text{Possible rational zeros} = \frac{\text{Factors of } 24}{\text{Factors of } 1} = \frac{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24}{\pm 1}
\]

Simplify each fraction:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24
\]

Final Answers:

1. Possible rational zeros: \( \boldsymbol{\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}} \)
2. Possible rational zeros: \( \boldsymbol{\pm 1, \pm 3, \pm 5, \pm 15} \)
3. Possible rational zeros: \( \boldsymbol{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24} \)