Explanation Procedure used: Procedure for Synthetic Division: “1. Write the coefficients of f ( x ) . Be certain that the powers are in descending order and that zeros are inserted for missing powers. 2. Carry down the left coefficient, then multiply it by r , and place this product under the second coefficient of the top line. 3. Add the two numbers in the second column and place the result below. Multiply this sum by r and place the product under the third coefficient of the top line. 4. Continue this process until the bottom row has as many numbers as the top row”. Complex Conjugate Root Theorem: “If the coefficients of the equation f ( x ) = 0 and real and a + b j ( b ≠ 0 ) is a complex root, then its conjugate, a − b j , is also a root”. Calculation: It is given that r 1 = 1 2 and r 2 = 1 + j are the roots of the equation V 5 + 4 V 4 + 5 V 3 − V 2 − 4 V − 5 = 0 . By complex conjugate root theorem, r 3 = 1 − j also root of the equation 2 x 6 − x 5 + 8 x 2 − 4 x = 0 . The equation can 2 x 6 − x 5 + 8 x 2 − 4 x = 0 be written as x ( 2 x 5 − x 4 + 8 x − 4 ) = 0 . The powers of x are in descending order, write down the coefficient of f ( x ) = 2 x 5 − x 4 + 8 x − 4 and zeros are inserted for missing powers and then place 1 to the right end. 2 − 1 0 0 8 − 4 1 2 Bring down the left coefficient 2 to the bottom line and multiply it by 1 2 , and place the product value 1 under the second coefficient –1 of the top line. Proceed in this similar manner, to find the quotient and remainder. 2 − 1 0 0 8 − 4 1 1 0 0 0 4 _ 2 0 0 0 8 0 1 + j [ ∵ r 2 = 1 + j also root ] 2 + 2 j 4 j − 4 + 4 j − 8 _ 2 2 + 2 j 4 j 4 + 4 j 0 1 − j [ ∵ r 3 = 1 − j also root ] 2 − 2 j 4 − 4 j 4 − 4 j _ 2 4 4 0 The numbers presented in the last row denotes the coefficients of the quotient and the remainder

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 15, Problem 36RE

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The roots of the equation 2x6−x5+8x2−4x=0 using synthetic division.

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Chapter 15, Problem 35RE

Chapter 15, Problem 37RE

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Task number: A1.1, A1.7 Topic: Celestial Navigation, Compass - Magnetic and Gyro Activ Determine compass error (magnetic and gyro) using azimuth choosing a suitable celestial body (Sun/ Stars/ Planets/ Moon). Apply variation to find the deviation of the magnetic compass. Minimum number of times that activity should be recorded: 6 (2 each phase) Sample calculation (Azimuth- Planets): On 06th May 2006 at 22h20m 10s UTC, a vessel in position 48°00'N 050°00'E observed Mars bearing 327° by compass. Find the compass error. If variation was 4.0° East, calculate the deviation. GHA Mars (06d 22h): Increment (20m 10s): 089° 55.7' 005° 02.5' v (0.9): (+) 00.3' GHA Mars: 094° 58.5' Longitude (E): (+) 050° 00.0' (plus- since longitude is easterly) LHA Mars: 144° 58.5' Declination (06d 22h): d (0.2): N 024° 18.6' (-) 00.1' Declination Mars: N 024° 18.5' P=144° 58.5' (If LHA<180°, P=LHA) A Tan Latitude/ Tan P A Tan 48° 00' Tan 144° 58.5' A = 1.584646985 N (A is named opposite to latitude, except when…

Chapter 15 Solutions

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 15.1 - Use the factor theorem to determine whether t − 2...Ch. 15.1 - Prob. 2PE Ch. 15.1 - Prob. 3PE Ch. 15.1 - Prob. 4PE Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - In Exercises 1–4, make the given changes in the...Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E Ch. 15.1 - Prob. 6E

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The roots of the equation 2 x 6 − x 5 + 8 x 2 − 4 x = 0 using synthetic division.

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Chapter 15 Solutions