1. Let F be a field and F be a subfield of E. The mapping : a(f)=f(a) is called the evaluation Given the evaluation homomorphism : C[x]→C, evaluate the given polynomial below. Note that (0) (0)=i^2=-1. F[x]→E defined by homomorphism. (2x³x² + 3x − 2) where i² = -1 - 2. Let F be a field and F be a subfield of E. The mapping : F[x]→Edefined by a(f)=f(a) is called the evaluation homomorphism. Given evaluation the homomorphism : Z_7[x]→Z_7, evaluate the given polynomial below. Note that is homomorphism and Z_7={0,1,2,3,4,5,6}. 3((x² + 2x)(x³ − 3x² + 3))
1. Let F be a field and F be a subfield of E. The mapping : a(f)=f(a) is called the evaluation Given the evaluation homomorphism : C[x]→C, evaluate the given polynomial below. Note that (0) (0)=i^2=-1. F[x]→E defined by homomorphism. (2x³x² + 3x − 2) where i² = -1 - 2. Let F be a field and F be a subfield of E. The mapping : F[x]→Edefined by a(f)=f(a) is called the evaluation homomorphism. Given evaluation the homomorphism : Z_7[x]→Z_7, evaluate the given polynomial below. Note that is homomorphism and Z_7={0,1,2,3,4,5,6}. 3((x² + 2x)(x³ − 3x² + 3))
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Polynomial And Rational Functions
Section9.2: Remainder And Factor Theorems
Problem 61PS
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Pls. solve 1 and 2. Thank you.
![1. Let F be a field and F be a subfield of E. The mapping :
a(f)=f(a) is called the evaluation
Given the evaluation homomorphism :
C[x]→C, evaluate the given polynomial below. Note that
(0) (0)=i^2=-1.
F[x]→E defined by
homomorphism.
(2x³x² + 3x − 2) where i² = -1
-
2. Let F be a field and F be a subfield of E. The mapping :
F[x]→Edefined by a(f)=f(a) is called the evaluation
homomorphism. Given
evaluation
the
homomorphism
: Z_7[x]→Z_7, evaluate the given
polynomial below. Note that is homomorphism and
Z_7={0,1,2,3,4,5,6}.
$3((x² + 2x)(x³ − 3x² + 3))](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdba100d8-ee81-4676-91a4-e53aa01f96c0%2Fbee6a067-3db5-41e2-ac5c-4b5ec6af23a0%2Fsv81dw8_processed.png&w=3840&q=75)
Transcribed Image Text:1. Let F be a field and F be a subfield of E. The mapping :
a(f)=f(a) is called the evaluation
Given the evaluation homomorphism :
C[x]→C, evaluate the given polynomial below. Note that
(0) (0)=i^2=-1.
F[x]→E defined by
homomorphism.
(2x³x² + 3x − 2) where i² = -1
-
2. Let F be a field and F be a subfield of E. The mapping :
F[x]→Edefined by a(f)=f(a) is called the evaluation
homomorphism. Given
evaluation
the
homomorphism
: Z_7[x]→Z_7, evaluate the given
polynomial below. Note that is homomorphism and
Z_7={0,1,2,3,4,5,6}.
$3((x² + 2x)(x³ − 3x² + 3))
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