Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 35: 141

Answer

(a) $$(A-1)(A+1) = A^2+A-A-1 = A^2-1$$ $$(A-1)(A^2+A+1) = A^3+A^2+A-A^2-A-1= A^3-1$$ $$(A-1)(A^3+A^2+A+1)= A^4+A^3+A^2+A -A^3-A^2-A-1 =A^4-1 $$ (b) $$A^5-1= (A-1)(A^4+A^3+A^2+A+1) = A^5+A^4+A^3+A^2+A-A^4-A^3-A^2-A-1 = A^5-1$$ $$A^n-1 = (A-1)(..(n-1)\times..+A^{n-3}+A^{n-2}+A^{n-1}+1)$$

Work Step by Step

(a) We will simply multiply right-hand side of these expressions and simplify them as possible: $(A-1)(A+1) = A^2+A-A-1 = A^2-1$ $(A-1)(A^2+A+1) = A^3+A^2+A-A^2-A-1= A^3-1$ $(A-1)(A^3+A^2+A+1)= A^4+A^3+A^2+A -A^3-A^2-A-1 =A^4-1 $ (b) According to the sequence we got in (a), we can consider that: $A^5-1$ will factor as $(A-1)(A^4+A^3+A^2+A+1)$ To verify, let's multiply and simplify this expression: $=> A^5+A^4+A^3+A^2+A-A^4-A^3-A^2-A-1 = A^5-1$ Now, according to the pattern we faced, we can generalize a factoring formula: $$A^n-1 = (A-1)(..(n-1)\times..+A^{n-3}+A^{n-2}+A^{n-1}+1)$$
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