Finding real zeroes.k(t) = 3t^6 - 54t^4 + 96t^2Please show all steps. This is an example of how it is done. **Educational Content on Polynomial Factorization and Solving for Roots**---**Problem 3: Polynomial Equation \( h(x) = x^4 - 12x^2 + 27 \)**1. **Equation Setup:** - \( 0 = x^4 - 12x^2 + 27 \)2. **Factorization:** - Identifying the greatest common factor (GCF): None (GCF = 1) - Factoring by substitution: - Let \( u = x^2 \) - Substitute \( u \) into the equation: \( u^2 - 12u + 27 = 0 \)3. **Quadratic Formula Application:** - Solving \( u^2 - 12u + 27 \) using the quadratic formula: - \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) - \( a = 1, b = -12, c = 27 \) - Solutions: \( u = 9 \) and \( u = 3 \)4. **Back Substitution:** - Replacing back to \( x^2 \): - \( x^2 = 9 \) and \( x^2 = 3 \) - Solving for real roots: - \( x = \pm 3 \), \( x = \pm \sqrt{3} \)5. **Real and Imaginary Roots:** - Result: 4 roots, 2 real (±3) and 2 imaginary (±√3)**Problem 4: Polynomial Equation \( j(x) = 4x^3 - 4x \)**1. **Equation Setup:** - \( 0 = 4x^3 - 4x \)2. **Factorization:** - Identifying the greatest common factor (GCF): \( 4x \) - Factoring: - \( 0 = 4x(x^2 - 1) \) - Difference of squares: \( (x^2 - 1) = (x-1)(x+1) \)3. **Roots Identification:** - Solving: \( 4x = 0 \), \( x-1 = 0 \), \(

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Finding real zeroes. k(t) = 3t^6 - 54t^4 + 96t^2 Please show all steps. This is an example of how it is done.

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Finding real zeroes.

k(t) = 3t^6 - 54t^4 + 96t^2

Please show all steps. This is an example of how it is done.

**Educational Content on Polynomial Factorization and Solving for Roots**

---

**Problem 3: Polynomial Equation \( h(x) = x^4 - 12x^2 + 27 \)**

1. **Equation Setup:**
- \( 0 = x^4 - 12x^2 + 27 \)

2. **Factorization:**
- Identifying the greatest common factor (GCF): None (GCF = 1)
- Factoring by substitution:
- Let \( u = x^2 \)
- Substitute \( u \) into the equation: \( u^2 - 12u + 27 = 0 \)

3. **Quadratic Formula Application:**
- Solving \( u^2 - 12u + 27 \) using the quadratic formula:
- \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( a = 1, b = -12, c = 27 \)
- Solutions: \( u = 9 \) and \( u = 3 \)

4. **Back Substitution:**
- Replacing back to \( x^2 \):
- \( x^2 = 9 \) and \( x^2 = 3 \)
- Solving for real roots:
- \( x = \pm 3 \), \( x = \pm \sqrt{3} \)

5. **Real and Imaginary Roots:**
- Result: 4 roots, 2 real (±3) and 2 imaginary (±√3)

**Problem 4: Polynomial Equation \( j(x) = 4x^3 - 4x \)**

1. **Equation Setup:**
- \( 0 = 4x^3 - 4x \)

2. **Factorization:**
- Identifying the greatest common factor (GCF): \( 4x \)
- Factoring:
- \( 0 = 4x(x^2 - 1) \)
- Difference of squares: \( (x^2 - 1) = (x-1)(x+1) \)

3. **Roots Identification:**
- Solving: \( 4x = 0 \), \( x-1 = 0 \), \(

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