## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set: 9

#### Answer

$(2t+3)(4t^2-6t+9)$

#### Work Step by Step

The expressions $8t^3$ and $27$ are both perfect cubes (the cube root is exact). Hence, $8t^3+27$ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2t)^3+(3)^3 \\\\= (2t+3)[(2t)^2-2t(3)+(3)^2] \\\\= (2t+3)(4t^2-6t+9) .\end{array}

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