## Precalculus (6th Edition)

Published by Pearson

# Chapter R - Review of Basic Concepts - Test: 11

#### Answer

$\color{blue}{3t^3+5t^2+2t+8}$

#### Work Step by Step

Distribute each term of the binomial factor to obtain: $=t(3t^2-t+4)+2(3t^2-t+4$) Distribute $t$ and $2$ to obtain: $=t(3t^2)-t(t)+t(4) +2(3t^2)-2(t)+2(4) \\=3t^3-t^2+4t+6t^2-2t+8$ Combine like terms to obtain: $=3t^3+(-t^2+6t^2)+(4t-2t)+8 \\=\color{blue}{3t^3+5t^2+2t+8}$

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