## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - Chapter Review Exercises: 60

#### Answer

(t+2)(t$^2$-2t+t$^2$)(t-2)(t$^2$+2t+t$^2$)

#### Work Step by Step

t$^6$-64 This expression is a difference of squares. If a is the square root of the first term and b is the square root of the second term, the factorization is (a+b)(a-b) $\sqrt {t^6}$=t$^3$ and $\sqrt {64}$=8. So the factorization is (t$^3$+8)(t$^3$-8). Now these two polynomials are a sum of cubes and a difference of cubes. If the cube root of the first term in each polynomial is a and the cube root of the second term in each polynomial is b, then the full factorization of both is (a+b)(a$^2$-ab+b$^2$)(a-b)(a$^2$+ab+b$^2$) $\sqrt[3] {t^3}$=t and $\sqrt[3] {8}$=2. So the factorization is (t+2)(t$^2$-2t+t$^2$)(t-2)(t$^2$+2t+t$^2$).

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