Example 6: Consider (x³ −7) € Q[x]. Let ae C be a root of (x³ -7) (in C). Find (1+ a)-¹. Q[x] Solution: {a+bx+cx²/a,b,ce Q1. request explain this step? Now a³ = 7. Also, since (1+x,x³ −7) = 1, by the Euclidean algorithm 3 f(x), g(x) = Q[x] such that (1+x)f(x) + (x³-7)g(x)=1, ~ In fact, here you can explicitly find f(x)=(x²- Now, putting x = α in (3), we get (1+x)f(x) = 1 )=1/(a²- i.e., (1+α)¹ = f(a)==(a²-a +1). 8 x²-x+1 xH) 2²-7 43 213 +x² -x2. -712 7 ан Quotient =x²-xH Remainder = -8 _n (x²-x+1) and g(x) n. 717 ...(3) How do we get this step o b(x) = f (x²-xH) 4. g(x) = ~{?
Example 6: Consider (x³ −7) € Q[x]. Let ae C be a root of (x³ -7) (in C). Find (1+ a)-¹. Q[x] Solution: {a+bx+cx²/a,b,ce Q1. request explain this step? Now a³ = 7. Also, since (1+x,x³ −7) = 1, by the Euclidean algorithm 3 f(x), g(x) = Q[x] such that (1+x)f(x) + (x³-7)g(x)=1, ~ In fact, here you can explicitly find f(x)=(x²- Now, putting x = α in (3), we get (1+x)f(x) = 1 )=1/(a²- i.e., (1+α)¹ = f(a)==(a²-a +1). 8 x²-x+1 xH) 2²-7 43 213 +x² -x2. -712 7 ан Quotient =x²-xH Remainder = -8 _n (x²-x+1) and g(x) n. 717 ...(3) How do we get this step o b(x) = f (x²-xH) 4. g(x) = ~{?
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 5T
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Request explain last step as to how do we get g(x)= -1/8 from the remainder and f(x)=-(1/8)*(x^2-x+1)
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