## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 7 - Cumulative Review - Page 564: 34

#### Answer

The answer is $4x^2 + 16x + 55 + \frac{222}{x - 4}$.

#### Work Step by Step

Now, the dividend $4x^3 - 9x + 2$ can be rewritten as $4x^3 + 0x^2 - 9x + 2$. $\space \space \space \space \space \space \space \space \space \space \space \space 4x^2 + 16x + 55$ $x - 4 /\overline{4x^3 + 0x^2 - 9x + 2}$ $\space \space \space \space \space \space \space \space \space \space \space \space \underline{4x^3 - 16x^2}$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 16x^2 - 9x$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \underline{16x^2 -64x}$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 55x + 2$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \underline{55x - 220}$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 222$ The answer is $4x^2 + 16x + 55 + \frac{222}{x - 4}$.

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