Chapter 11.3, Problem 34PPE

To determine

To divide $\left(7b+16b^3\right)÷\left(-1+8b\right)$ .

Answer to Problem 34PPE

Dividing $\left(7b+16b^3\right)÷\left(-1+8b\right)$ the quotient is $2b^2+\frac{1}{4}b+\frac{29}{32}$ and remainder is $\frac{29}{32}$ .

Explanation of Solution

Given:

  $\left(7b+16b^3\right)÷\left(-1+8b\right)$

Concept Used:

Steps to follow before applying Long Division method:

1. Write the terms of the dividend and divisor in descending powers of the variable.
2. Insert placeholders with zero coefficients for missing powers of the variable.

Calculation:

To divide $\left(7b+16b^3\right)÷\left(-1+8b\right)$ use long division method.

First, arrange the terms of the dividend and divisor in descending powers of the variable, as

  $\left(16b^3+7b\right)÷\left(8b-1\right)$

Now, insert placeholders with zero coefficients for missing powers of the variable, as

  $\left(16b^3+0b^2+7b\right)÷\left(8b-1\right)$

Thus, the given division may be proceeded as:

  $\begin{array}{l}8b-1)16b^3+0b^2+7b(2b^2+\frac{1}{4}b+\frac{29}{32}\\ \underset{\_}{16b^3-2b^2}\\ \underset{\_}{\begin{array}{l}2b^2+7b\\ 2b^2-\frac{1}{4}b\end{array}}\\ \left(7+\frac{1}{4}\right)b=\frac{29}{4}b\\ \underset{\_}{\frac{29}{4}b-\frac{29}{32}}\\ \frac{29}{32}\end{array}$

Thus, dividing $\left(7b+16b^3\right)÷\left(-1+8b\right)$ the quotient is $2b^2+\frac{1}{4}b+\frac{29}{32}$ and remainder is $\frac{29}{32}$ .