Lesson 3 Solutions

# Lesson 3 Solutions - Problem Set Factor each of these...

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Factor each of these polynomials: 1. f ( x ) = 2 x 2 + 3 x - 2 By Descartes Rule ( DR ), there is one positive and one negative root. Possible roots are: 1, 2, 1 2 , -1, -2, - 1 2 f (1) = 3 f (2) = 5 f ( 1 2 ) = 0 So 1 2 is one of the roots. By Vieta’s Formula ( V F ), - 3 2 is the sum of both roots If x is the other root: - 3 2 = 1 2 + x . Then x = - 2. Therefore f ( x ) = (2 x - 1)( x + 2) 2. f ( x ) = 6 x 2 + x - 2 By ( DR ), there is one positive and one negative root. Possible roots are: 1, 2, 1 2 , 1 3 , 2 3 , -1, -2, - 1 2 , - 1 3 , - 2 3 f (1) = 5 f (2) = 24 f ( 1 2 ) = 0 So 1 2 is one of the roots. By ( V F ), - 1 6 is the sum of both roots. If x is the other root: - 1 6 = 1 2 + x . Then x = - 2 3 . Therefore f ( x ) = (2 x - 1)(3 x + 2) 3. f ( x ) = 2 x 2 + 7 x + 3 By ( DR ), there are no positive and two (or zero) negative roots. Possible roots are: -1, -3, - 1 2 , - 3 2 f ( - 1) = - 2 f ( - 3) = 0 So -3 is one of the roots. By (

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## This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton.

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Lesson 3 Solutions - Problem Set Factor each of these...

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