## Algebra 1

Published by Prentice Hall

# Chapter 8 - Polynomials and Factoring - 8-8 Factoring by Grouping - Practice and Problem-Solving Exercises: 46

#### Answer

$(2^{5} + 2^{4}) + (2^{3} + (2^{2}) + (2^{1} + 2^{0})$ = (21)(3) = 63

#### Work Step by Step

$(2^{5} + 2^{4}) + (2^{3} + (2^{2}) + (2^{1} + 2^{0})$ We factor the GCF from the first two, middle two and the last three terms $2^{4}(2^{1} + 2^{0}) + 2^{2}(2^{1} + 2^{0}) + 1(2^{1} + 2^{0})$ We factor out the common $(2^{1} + 2^{0})$ and we get $(2^{4} + 2^{2} + 1)(2^{1} + 2^{0})$ We simplify the brackets $2^{4}$ = 2*2*2*2 = 16 $2^{3}$ = 2*2*2 = 8 $2^{2}$ = 2*2 = 4 $2^{1}$ = 2 $2^{0}$ = 1 (16+4+1) (2+1) (21)(3) 63

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.