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Theorems About Roots of Polynomial Equations

Theorems About Roots of Polynomial Equations. ALGEBRA 2 LESSON 6-5. Algebra 2 Lesson 6-5 (Page 329). 6-5. Theorems About Roots of Polynomial Equations. ALGEBRA 2 LESSON 6-5. Rational Root Theorem. Represents the simplest form of a root of a polynomial equation.

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Theorems About Roots of Polynomial Equations

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  1. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 Algebra 2 Lesson 6-5 (Page 329) 6-5

  2. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 Rational Root Theorem Represents the simplest form of a root of a polynomial equation p = factors of the constant term q = factors of the leading coefficient 6-5

  3. 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, ± . 5 3 1 3 Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 Find the rational roots of 3x3 – x2 – 15x + 5. Step 1: List the possible rational roots. 6-5

  4. / 5: 3(5)3 – (5)2 – 15(5) + 5 = 280 = 0 –5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 = 0 / 5 3 / : 3 3 – 2 – 15 + 5 = –8.8 = 0 : 3 3 – 2 – 15 + 5 = –13.3 = 0 5 3 / – 1 3 5 3 ( ) ( ) 1 3 1 3 5 3 5 3 ( ) ( ) ( ) ( ) 1: 3(1)3 – (1)2 – 15(1) + 5 = –8 = 0 –1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 = 0 / / 1 3 1 3 : 3 3 – 2 – 15 + 5 = 0 So is a root. : 3 3 – 2 – 15 + 5 = 9.7 = 0 5 3 5 3 5 3 ( ) ( ) ( ) – – – 1 3 1 3 ( ) 1 3 ( ) 1 3 ( ) / – – – – 1 3 The only rational root of 3x3 – x2 – 15x + 5 = 0 is . Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 (continued) Step 2: Test each possible rational root. 6-5

  5. 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. 1 5 2 5 4 5 Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 Find the roots of 5x3 – 24x2 + 41x – 20 = 0. Step 1: List the possible rational roots. 6-5

  6. 1 5 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. / 1 5 / 2 5 / 2 5 1 5 1 5 1 5 2 5 4 5 4 5 4 5 2 5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 5 / 4 5 4 5 4 5 5 –24 41 –20 4 –16 20 5 –20 25 0 5x2 – 20x + 25  Remainder 1 5 1 5 1 5 2 5 2 5 2 5 (– ) (– ) (– ) (– ) (– ) (– ) Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 (continued) 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. 6-5

  7. –b ± b2 – 4ac 2a x = Quadratic Formula = Substitute 1 for a, –4 for b, and 5 for c. –(–4) ± (–4)2 – 4(1)(5) 2(1) = Use order of operations. =    –1 = i. = 2 ± iSimplify. 4 ± –4 2 4 ± 2i 2 4 5 The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 (continued) Step 4: Find the roots of 5x2 – 20x + 25 = 0. 5x2 – 20x + 25 = 0 5(x2 – 4x + 5) = 0 Factor out the GCF, 5. x2 – 4x + 5 = 0 6-5

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

  9. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 A polynomial equation and real coefficients has the roots 2 + 9i with 7i. Find two additional roots. 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. 6-5

  10. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i. 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)) 6-5

  11. Theorems About Roots of Polynomial Equations ALGEBRA 2 LESSON 6-5 (continued) Step 3: Multiply the factors. (x + 2)[x2 – x(2 – i) – x(2 + i) + (2 – i)(2 + i)] Multiply (x – (2 – i)) (x – (2 + i)). (x + 2)(x2 – 2x + ix – 2x – ix + 4 – i2) Simplify. (x + 2)(x2 – 2x – 2x + 4 + 1) (x + 2)(x2 – 4x + 5) Multiply. x3 – 2x2 – 3x + 10 A third-degree polynomial equation with rational coefficients and roots –2 and 2 – i is x3 – 2x2 – 3x + 10 = 0. 6-5

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