1 / 12

6-5 Theorems About Roots of Polynomial Equations

6-5 Theorems About Roots of Polynomial Equations. Objectives. The Rational Root Theorem Irrational Root Theorem & Imaginary Root Theorem. The leading coefficient is 3. The constant term is 5. By the Rational Root Theorem, the only possible rational roots of the equation have

yazid
Télécharger la présentation

6-5 Theorems About Roots of Polynomial Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 6-5 Theorems About Roots of Polynomial Equations

  2. Objectives The Rational Root Theorem Irrational Root Theorem & Imaginary Root Theorem

  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 Finding Rational Roots Find the rational roots of 3x3 – x2 – 15x + 5. Step 1: List the possible rational roots.

  4. 5 3 : 3 3 – 2 – 15 + 5 = –8.8 ≠ 0 : 3 3 – 2 – 15 + 5 = –13.3 ≠ 0 5 3 – 1 3 5 3 ( ) ( ) 5 3 1 3 1 3 5 3 ( ) ( ) ( ) ( ) 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 . Continued (continued) Step 2: Test each possible rational root. 5: 3(5)3 – (5)2 – 15(5) + 5 = 280 ≠ 0 –5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 ≠ 0 1: 3(1)3 – (1)2 – 15(1) + 5 = –8 ≠ 0 –1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 ≠ 0

  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 Using the Rational Root Theorem Find the roots of 5x3 – 24x2 + 41x – 20 = 0. Step 1: List the possible rational roots.

  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 2 5 4 5 4 5 2 5 1 5 2 5 4 5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 5 4 5 4 5 5 –24 41 –20 4 –16 20 5 –20 25 0 5x2 – 20x + 25  Remainder 2 5 1 5 1 5 2 5 2 5 1 5 (– ) (– ) (– ) (– ) (– ) (– ) Continued (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.

  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. Continued (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

  8. A polynomial equation with rational coefficients has the roots2 – 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. Finding Irrational Roots

  9. Finding Imaginary Roots 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.

  10. Writing a Polynomial Equation from Its Roots 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))

  11. Continued (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.

  12. Homework Pg 339 #1, 7, 13, 14, 15, 19, 20

More Related