This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Assignment 3 Solutions Each question is worth 1 point unless otherwise marked. Marked out of 10 Factor each of these polynomials: 1. f ( x ) = 2 x 2 x 1 f (1) = 0. By Vieta’s formula, 1/2 is the sum of the roots. r 1 + r 2 = 1 / 2 1 + r 2 = 1 / 2 r 2 = 1 / 2 f ( x ) = ( x 1)(2 x + 1) = 2( x 1)( x 1 / 2) 2. f ( x ) = 5 x 2 + 5 x 10 f ( x ) = 5( x 2 + x 2) f (1) = 0. By Vieta’s formula, 1 is the sum of the roots. r 1 + r 2 = 1 1 + r 2 = 1 r 2 = 2 f ( x ) = 5( x 1)( x + 2) 3. f ( x ) = 2 x 2 5 x 3 f (1) = 6, f ( 1) = 4, f (3) = 0 By Vieta’s formula, 5/2 is the sum of the roots. r 1 + r 2 = 5 / 2 3 + r 2 = 5 / 2 r 2 = 1 / 2 f ( x ) = ( x 3)(2 x + 1) = 2( x 3)( x + 1 / 2) 4. f ( x ) = 1 2 x 2 + 2 x + 3 2 f ( x ) = 1 2 ( x 2 + 4 x + 3) Since the coefficients are all positive, there are only negative roots. This means the only possible rational roots are 1 and 3. Since f (1) = 0 and f (3) = 0, we get the polynomial: f ( x ) = 1 2 ( x + 1)( x + 3) 5. A child’s age increased by 5 years gives a number which has a square root5....
View
Full
Document
This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton.
 Fall '10
 Unknown
 Linear Algebra, Algebra, Polynomials

Click to edit the document details