Assignment 3 Solutions

Assignment 3 Solutions - Assignment 3 Solutions Each...

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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....
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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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Assignment 3 Solutions - Assignment 3 Solutions Each...

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