11:09 PM Question 14 (()) 884 4G 4G B/S 47 Part 1 of 3 PART I In this exercise we will use synthetic division to factor the polynomial f(x) = x³-21x- 20. This will start as a division problem, where you are given two factors to "pull out" of f(x): 3 21x 20 (x + 1)(x-5) You'll notice that the denominator is a quadratic polynomial, but synthetic division only works on a single factor at a time. This means you must perform synthetic division on each factor separately. It goes quick once you get the hang of it. It does not matter which factor on the denominator you start with, so let's pick (x + 1), performing synthetic division the usual way. -1 1 0 -21 -20 -1 1 20 1 -1 -20 0 The quotient is: x-x-20 The remainder is: 0 PART II Great! Your remainder of zero proves that (x + 1) was a factor. We can now write this: 3 21x 20 (x²-x-20)(x+1) x² -20 (x + 1)(x-5) (x + 1)(x-5) (x-5) 0.56 Part 2 of 3 Time to divide by that last factor of (x-5). But first, a public service announcement: WARNING: When using synthetic division to factor a polynomial, a non-zero remainder means STOP! You are not working with a factor, and you need to use long division if you want to go any further! We are ok to keep going since the remainder was zero and (x + 1) was a factor. Use the previous quotient as the top line of a new synthetic division table. x2-x-20 (x-5) 1 -21 -20 -1 1 20 1 -1 -20 0 The quotient is: The remainder is: Question Help: Video Post to forum Submit Question ← ||| → 51 "
11:09 PM Question 14 (()) 884 4G 4G B/S 47 Part 1 of 3 PART I In this exercise we will use synthetic division to factor the polynomial f(x) = x³-21x- 20. This will start as a division problem, where you are given two factors to "pull out" of f(x): 3 21x 20 (x + 1)(x-5) You'll notice that the denominator is a quadratic polynomial, but synthetic division only works on a single factor at a time. This means you must perform synthetic division on each factor separately. It goes quick once you get the hang of it. It does not matter which factor on the denominator you start with, so let's pick (x + 1), performing synthetic division the usual way. -1 1 0 -21 -20 -1 1 20 1 -1 -20 0 The quotient is: x-x-20 The remainder is: 0 PART II Great! Your remainder of zero proves that (x + 1) was a factor. We can now write this: 3 21x 20 (x²-x-20)(x+1) x² -20 (x + 1)(x-5) (x + 1)(x-5) (x-5) 0.56 Part 2 of 3 Time to divide by that last factor of (x-5). But first, a public service announcement: WARNING: When using synthetic division to factor a polynomial, a non-zero remainder means STOP! You are not working with a factor, and you need to use long division if you want to go any further! We are ok to keep going since the remainder was zero and (x + 1) was a factor. Use the previous quotient as the top line of a new synthetic division table. x2-x-20 (x-5) 1 -21 -20 -1 1 20 1 -1 -20 0 The quotient is: The remainder is: Question Help: Video Post to forum Submit Question ← ||| → 51 "
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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