1. Algebra 2 List all the integer factors for the number below: 36

2. Algebra 2

3. A. the set of numbers that can be written in the form a + bi, where i represents the square root of -1 B. the set of numbers that can be written as a quotient of integers C. numbers that cannot be written as quotients of integers D. the set of natural numbers, their opposites, and zero What are the rational numbers? What are the irrational numbers? What are the imaginary numbers? What does the remaining option describe?

4. Algebra 2 Rational Root theorem

5. Algebra 2 Find the rational roots of 3x 3 – x 2 – 15x + 5. Lesson 6-5 Theorems About Roots of Polynomial Equations Step 1: List the possible rational roots. The leading coefficient is 3. The constant term is 5. By the Rational Root Theorem, the only possible rational roots of the equation have the form. factors of 5 factors of 3 The factors of 5 are ±5 and ±1 and ±5. The factors of 3 are ±3 and ±1. The only possible rational roots are ±5, ±, ±1, ±. 5353 1313 Additional Examples

6. Algebra 2 (continued) Lesson 6-5 Theorems About Roots of Polynomial Equations Step 2:Test each possible rational root. 1: 3(1) 3 – (1) 2 – 15(1) + 5 = –8 ≠ 0 –1: 3(–1) 3 – (–1) 2 – 15(–1) + 5 = 16 ≠ 0 5: 3(5) 3 – (5) 2 – 15(5) + 5 = 280 ≠ 0 –5: 3(–5) 3 – (–5) 2 – 15(–5) + 5 = –320 ≠ 0 : 3 3 – 2 – 15 + 5 = –8.8 ≠ 0 : 3 3 – 2 – 15 + 5 = –13.3 ≠ 0 5353 5353 ( ) 5353 5353 5353 – 5353 – 5353 – 5353 – : 3 3 – 2 – 15 + 5 = 0 So is a root. : 3 3 – 2 – 15 + 5 = 9.7 ≠ 0 1313 ( ) 1313 – 1313 – 1313 – 1313 1313 1313 1313 1313 – The only rational root of 3x 3 – x 2 – 15x + 5 = 0 is. 1313 Additional Examples

7. Algebra 2 Find the roots of 5x 3 – 24x 2 + 41x – 20 = 0. Lesson 6-5 Theorems About Roots of Polynomial Equations Step 1: List the possible rational roots. The leading coefficient is 5. The constant term is 20. By the Rational Root Theorem, the only possible roots of the equation have the form. factors of – 20 factors of 5 The factors of –20 are ±1 and ±20, ±2 and ±10, and ±4 and ±5. The only factors of 5 are ±1 and ±5. The only possible rational roots are ±, ±, ±, ±1, ±2, ±4, ±5, ±10, and ±20. 1515 2525 4545 Additional Examples

8. Algebra 2 (continued) Lesson 6-5 Theorems About Roots of Polynomial Equations Step 2: Test each possible rational root until you find a root. Step 3: Use synthetic division with the root you found in Step 2 to find the quotient. 5–2441–20 4 –16 20 5–2025 0 5x 2 – 20x + 25 Remainder 4545 4545 Test : 5 3 – 24 2 ± 41 – 20 = –12.72 ≠ 0 Test – : 5 3 – 24 2 ± 41 – 2 = –29.2 ≠ 0 Test : 5 3 – 24 2 ± 41 – 20 = –7.12 ≠ 0 Test – : 5 3 – 24 2 ± 41 – 20 = –40.56 ≠ 0 Test : 5 3 – 24 2 ± 41 – 20 = 0 So is a root. ( ) 1515 1515 1515 2525 2525 4545 2525 4545 1515 2525 4545 4545 2525 1515 ( – ) 1515 1515 1515 2525 2525 2525 Additional Examples

9. Algebra 2 (continued) Lesson 6-5 Theorems About Roots of Polynomial Equations Step 4:Find the roots of 5x 2 – 20x + 25 = 0. 5x 2 – 20x + 25 = 0 5(x 2 – 4x + 5) = 0Factor out the GCF, 5. x 2 – 4x + 5 = 0 x = Quadratic Formula = Substitute 1 for a, –4 for b, and 5 for c. –b ± b 2 – 4ac 2a –(–4) ± (–4) 2 – 4(1)(5) 2(1) = Use order of operations. = –1 = i. = 2 ± iSimplify. 4 ± –4 2 4 ± 2i 2 The roots of 5x 3 – 24x 2 + 41x – 20 = 0 are, 2 + i, and 2 – i. 4545 Additional Examples

10. Algebra 2 Homework: Page 333 1-4 all

11. Algebra 2 Irrational Root Theorem

12. Algebra 2 By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate 2 + 5 is also a root. A polynomial equation with rational coefficients has the roots 2 – 5 and 7. Find two additional roots. Lesson 6-5 Theorems About Roots of Polynomial Equations If 7 is a root, then its conjugate – 7 also is a root. Additional Examples

13. Algebra 2 Imaginary Root Theorem

14. Algebra 2 A polynomial equation and real coefficients has the roots 2 + 9i with 7i. Find two additional roots. Lesson 6-5 Theorems About Roots of Polynomial Equations By the Imaginary Root Theorem, if 2 + 9i is a root, then its complex conjugate 2 – 9i also is a root. If 7i is a root, then its complex conjugate –7i also is a root. Additional Examples

15. Algebra 2 Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i. Lesson 6-5 Theorems About Roots of Polynomial Equations Step 1: Find the other root using the Imaginary Root Theorem. Since 2 – i is a root, then its complex conjugate 2 + i is a root. Step 2: Write the factored form of the polynomial using the Factor Theorem. (x + 2)(x – (2 – i))(x – (2 + i)) Additional Examples

16. Algebra 2 (continued) Lesson 6-5 Theorems About Roots of Polynomial Equations Step 3: Multiply the factors. (x + 2)[x 2 – x(2 – i) – x(2 + i) + (2 – i)(2 + i)]Multiply (x – (2 – i)) (x – (2 + i)). (x + 2)(x 2 – 2x + ix – 2x – ix + 4 – i 2 )Simplify. (x + 2)(x 2 – 2x – 2x + 4 + 1) (x + 2)(x 2 – 4x + 5)Multiply. x 3 – 2x 2 – 3x + 10 A third-degree polynomial equation with rational coefficients and roots –2 and 2 – i is x 3 – 2x 2 – 3x + 10 = 0. Additional Examples

17. Algebra 2 Homework: Page 333 13-23 odd