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 "
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter3: Polynomial And Rational Functions
Section: Chapter Questions
Problem 2P: Too Many Corn Plants per Acre? The more corn a farmer plants per acre, the greater is the yield the...
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Transcribed Image Text: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:
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