Precalculus worksheet focusing on the Fundamental Theorem of Algebra, with exercises to determine the number of zeros and factor polynomials.

Problem Analysis

The task involves analyzing polynomial functions using the Fundamental Theorem of Algebra and related concepts such as:
1. Number of complex zeros: A polynomial of degree \( n \) has exactly \( n \) complex zeros (counting multiplicities).
2. Possible number of real and imaginary zeros: Real zeros are those that are real numbers, while imaginary zeros are non-real complex numbers.
3. Possible number of positive and negative zeros: Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros.
4. Possible rational zeros: Use the Rational Root Theorem to list all possible rational zeros.
5. Factoring into linear and irreducible quadratic factors: Factor the polynomial completely over the real or complex numbers.

We will solve each part step by step.

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Part 1: Analyze the Polynomials

#### 1. \( f(x) = 5x^4 - 36x^2 - 81 \)

- Degree: 4
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 4 complex zeros (real or imaginary).

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 5x^4 - 36x^2 - 81 \).
- Signs: \( +, -, -, - \)
- Number of sign changes: 1
- Possible positive real zeros: 1

- Negative zeros: Count sign changes in \( f(-x) = 5(-x)^4 - 36(-x)^2 - 81 \).
- Simplify: \( f(-x) = 5x^4 - 36x^2 - 81 \)
- Signs: \( +, -, -, - \)
- Number of sign changes: 1
- Possible negative real zeros: 1

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (-81) and \( q \) divides the leading coefficient (5).
- \( p = \pm 1, \pm 3, \pm 9, \pm 27, \pm 81 \)
- \( q = \pm 1, \pm 5 \)
- Possible rational zeros: \( \pm 1, \pm 3, \pm 9, \pm 27, \pm 81, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{9}{5}, \pm \frac{27}{5}, \pm \frac{81}{5} \)

#### 2. \( f(x) = 15x^5 + 3x^4 + 140x^3 + 28x^2 + 45x + 9 \)

- Degree: 5
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 5 complex zeros.

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 15x^5 + 3x^4 + 140x^3 + 28x^2 + 45x + 9 \).
- Signs: \( +, +, +, +, +, + \)
- Number of sign changes: 0
- Possible positive real zeros: 0

- Negative zeros: Count sign changes in \( f(-x) = 15(-x)^5 + 3(-x)^4 + 140(-x)^3 + 28(-x)^2 + 45(-x) + 9 \).
- Simplify: \( f(-x) = -15x^5 + 3x^4 - 140x^3 + 28x^2 - 45x + 9 \)
- Signs: \( -, +, -, +, -, + \)
- Number of sign changes: 5
- Possible negative real zeros: 5, 3, or 1

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (9) and \( q \) divides the leading coefficient (15).
- \( p = \pm 1, \pm 3, \pm 9 \)
- \( q = \pm 1, \pm 3, \pm 5, \pm 15 \)
- Possible rational zeros: \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm \frac{3}{5} \)

#### 3. \( f(x) = 5x^3 - x^2 - 5x + 1 \)

- Degree: 3
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 3 complex zeros.

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 5x^3 - x^2 - 5x + 1 \).
- Signs: \( +, -, -, + \)
- Number of sign changes: 2
- Possible positive real zeros: 2 or 0

- Negative zeros: Count sign changes in \( f(-x) = 5(-x)^3 - (-x)^2 - 5(-x) + 1 \).
- Simplify: \( f(-x) = -5x^3 - x^2 + 5x + 1 \)
- Signs: \( -, -, +, + \)
- Number of sign changes: 1
- Possible negative real zeros: 1

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (1) and \( q \) divides the leading coefficient (5).
- \( p = \pm 1 \)
- \( q = \pm 1, \pm 5 \)
- Possible rational zeros: \( \pm 1, \pm \frac{1}{5} \)

#### 4. \( f(x) = 3x^3 + 11x^2 + 5x - 3 \)

- Degree: 3
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 3 complex zeros.

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 3x^3 + 11x^2 + 5x - 3 \).
- Signs: \( +, +, +, - \)
- Number of sign changes: 1
- Possible positive real zeros: 1

- Negative zeros: Count sign changes in \( f(-x) = 3(-x)^3 + 11(-x)^2 + 5(-x) - 3 \).
- Simplify: \( f(-x) = -3x^3 + 11x^2 - 5x - 3 \)
- Signs: \( -, +, -, - \)
- Number of sign changes: 2
- Possible negative real zeros: 2 or 0

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (-3) and \( q \) divides the leading coefficient (3).
- \( p = \pm 1, \pm 3 \)
- \( q = \pm 1, \pm 3 \)
- Possible rational zeros: \( \pm 1, \pm 3, \pm \frac{1}{3} \)

#### 5. \( f(x) = 10x^5 - 15x^4 + 12x^3 - 18x^2 + 2x - 3 \)

- Degree: 5
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 5 complex zeros.

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 10x^5 - 15x^4 + 12x^3 - 18x^2 + 2x - 3 \).
- Signs: \( +, -, +, -, +, - \)
- Number of sign changes: 5
- Possible positive real zeros: 5, 3, or 1

- Negative zeros: Count sign changes in \( f(-x) = 10(-x)^5 - 15(-x)^4 + 12(-x)^3 - 18(-x)^2 + 2(-x) - 3 \).
- Simplify: \( f(-x) = -10x^5 - 15x^4 - 12x^3 - 18x^2 - 2x - 3 \)
- Signs: \( -, -, -, -, -, - \)
- Number of sign changes: 0
- Possible negative real zeros: 0

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (-3) and \( q \) divides the leading coefficient (10).
- \( p = \pm 1, \pm 3 \)
- \( q = \pm 1, \pm 2, \pm 5, \pm 10 \)
- Possible rational zeros: \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{1}{10}, \pm \frac{3}{10} \)

#### 6. \( f(x) = 5x^5 - 25x^4 + 46x^3 - 230x^2 + 9x - 45 \)

- Degree: 5
- Complex Zeros: By the Fundamental Theorem of Algebra, there are 5 complex zeros.

Descartes' Rule of Signs:
- Positive zeros: Count sign changes in \( f(x) = 5x^5 - 25x^4 + 46x^3 - 230x^2 + 9x - 45 \).
- Signs: \( +, -, +, -, +, - \)
- Number of sign changes: 5
- Possible positive real zeros: 5, 3, or 1

- Negative zeros: Count sign changes in \( f(-x) = 5(-x)^5 - 25(-x)^4 + 46(-x)^3 - 230(-x)^2 + 9(-x) - 45 \).
- Simplify: \( f(-x) = -5x^5 - 25x^4 - 46x^3 - 230x^2 - 9x - 45 \)
- Signs: \( -, -, -, -, -, - \)
- Number of sign changes: 0
- Possible negative real zeros: 0

Rational Root Theorem:
- Possible rational zeros: \( \pm \frac{p}{q} \), where \( p \) divides the constant term (-45) and \( q \) divides the leading coefficient (5).
- \( p = \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45 \)
- \( q = \pm 1, \pm 5 \)
- Possible rational zeros: \( \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{9}{5}, \pm \frac{15}{5}, \pm \frac{45}{5} \)

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Part 2: Factor the Polynomials

#### 7. \( f(x) = 2x^4 - 11x^2 + 9 \)

This is a quartic polynomial, but it can be treated as a quadratic in \( x^2 \):
- Let \( u = x^2 \). Then \( f(x) = 2u^2 - 11u + 9 \).

Factor the quadratic:
\[ 2u^2 - 11u + 9 = (2u - 1)(u - 9) \]

Substitute back \( u = x^2 \):
\[ f(x) = (2x^2 - 1)(x^2 - 9) \]

Factor further:
\[ x^2 - 9 = (x - 3)(x + 3) \]

Thus:
\[ f(x) = (2x^2 - 1)(x - 3)(x + 3) \]

#### 8. \( f(x) = 27x^3 + 1 \)

This is a sum of cubes:
\[ 27x^3 + 1 = (3x)^3 + 1^3 \]

Use the sum of cubes formula:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
where \( a = 3x \) and \( b = 1 \):
\[ 27x^3 + 1 = (3x + 1)((3x)^2 - (3x)(1) + 1^2) \]
\[ = (3x + 1)(9x^2 - 3x + 1) \]

The quadratic \( 9x^2 - 3x + 1 \) has no real roots (discriminant \( b^2 - 4ac = (-3)^2 - 4(9)(1) = 9 - 36 = -27 < 0 \)), so it is irreducible over the reals.

Thus:
\[ f(x) = (3x + 1)(9x^2 - 3x + 1) \]

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Final Answers

1. \( f(x) = 5x^4 - 36x^2 - 81 \)
- Complex zeros: 4
- Real zeros: 2 (1 positive, 1 negative)
- Imaginary zeros: 2
- Rational zeros: Possible candidates are \( \pm 1, \pm 3, \pm 9, \pm 27, \pm 81, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{9}{5}, \pm \frac{27}{5}, \pm \frac{81}{5} \)

2. \( f(x) = 15x^5 + 3x^4 + 140x^3 + 28x^2 + 45x + 9 \)
- Complex zeros: 5
- Real zeros: 1 (negative)
- Imaginary zeros: 4
- Rational zeros: Possible candidates are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm \frac{3}{5} \)

3. \( f(x) = 5x^3 - x^2 - 5x + 1 \)
- Complex zeros: 3
- Real zeros: 1 (positive), 1 (negative)
- Imaginary zeros: 1
- Rational zeros: Possible candidates are \( \pm 1, \pm \frac{1}{5} \)

4. \( f(x) = 3x^3 + 11x^2 + 5x - 3 \)
- Complex zeros: 3
- Real zeros: 1 (positive), 2 (negative)
- Imaginary zeros: 0
- Rational zeros: Possible candidates are \( \pm 1, \pm 3, \pm \frac{1}{3} \)

5. \( f(x) = 10x^5 - 15x^4 + 12x^3 - 18x^2 + 2x - 3 \)
- Complex zeros: 5
- Real zeros: 1 (positive)
- Imaginary zeros: 4
- Rational zeros: Possible candidates are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{1}{10}, \pm \frac{3}{10} \)

6. \( f(x) = 5x^5 - 25x^4 + 46x^3 - 230x^2 + 9x - 45 \)
- Complex zeros: 5
- Real zeros: 1 (positive)
- Imaginary zeros: 4
- Rational zeros: Possible candidates are \( \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45, \pm \frac{1}{5}, \pm \frac{3}{5}, \pm \frac{9}{5}, \pm \frac{15}{5}, \pm \frac{45}{5} \)

7. \( f(x) = 2x^4 - 11x^2 + 9 \)
- Factored form: \( (2x^2 - 1)(x - 3)(x + 3) \)

8. \( f(x) = 27x^3 + 1 \)
- Factored form: \( (3x + 1)(9x^2 - 3x + 1) \)

\boxed{\text{See detailed solutions above}}