Determine to the 24 +22+8=0 Need (f(2)| > | g (2) on 121= 2 19 (2)| = 22 +8 ≤ 2/2/+8 = 2·2+8 12 If(z) | = |21ª = 16 | f(z) | > |g(z) | So 16 >12 T So of za f zeroes and = and ftg in 121<2 +22+8 4 Zeroes number with 2 4 on of solutions 12/<2 4 zeroes 121=2 counting multipitciy (use Rourke' on g(2)= 22 +8 have the same number f(2)=24
Determine to the 24 +22+8=0 Need (f(2)| > | g (2) on 121= 2 19 (2)| = 22 +8 ≤ 2/2/+8 = 2·2+8 12 If(z) | = |21ª = 16 | f(z) | > |g(z) | So 16 >12 T So of za f zeroes and = and ftg in 121<2 +22+8 4 Zeroes number with 2 4 on of solutions 12/<2 4 zeroes 121=2 counting multipitciy (use Rourke' on g(2)= 22 +8 have the same number f(2)=24
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 24E
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Can you please explain the solution (attached below) step by step. I don't understand this problem.
![Determine
to
the
24 +22+8=0
Need If(2)| > | g (2) on 121= 2
19 (2)| = 22 +8 ≤ 2/2/+8
= 2·2+8
12
If(z) | = |21ª = 16
| f(z) | > |g(z) |
So
16 >12
T
So
of
za
f
zeroes
and
=
and ftg
in 121<2
+22+8
4 Zeroes
number
with
2
4
on
of solutions
12/<2
4 zeroes
121=2
counting multipitciy
(use Rourke' on
g(2)= 22 +8
have the same
number
f(2)=24](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa62f7b35-7db6-46d8-92c3-a45ad2747ea7%2Fe5ed7df2-482e-48c1-98fb-8ef4d883de59%2Fvz1xddi_processed.png&w=3840&q=75)
Transcribed Image Text:Determine
to
the
24 +22+8=0
Need If(2)| > | g (2) on 121= 2
19 (2)| = 22 +8 ≤ 2/2/+8
= 2·2+8
12
If(z) | = |21ª = 16
| f(z) | > |g(z) |
So
16 >12
T
So
of
za
f
zeroes
and
=
and ftg
in 121<2
+22+8
4 Zeroes
number
with
2
4
on
of solutions
12/<2
4 zeroes
121=2
counting multipitciy
(use Rourke' on
g(2)= 22 +8
have the same
number
f(2)=24
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