Chapter 11.3, Problem 30PPE

To determine

To divide $\left(-13x+6x^3-6-x^2\right)÷\left(3x-5\right)$ .

Answer to Problem 30PPE

Dividing $\left(-13x+6x^3-6-x^2\right)÷\left(3x-5\right)$ the quotient is $2x^2+3x+\frac{2}{3}$ and remainder is $\frac{-8}{3}$ .

Explanation of Solution

Given:

  $\left(-13x+6x^3-6-x^2\right)÷\left(3x-5\right)$

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 $\left(-13x+6x^3-6-x^2\right)÷\left(3x-5\right)$ use long division method.

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

  $\left(6x^3-x^2-13x-6\right)÷\left(3x-5\right)$

The division may be proceeded as:

  $\begin{array}{l}3x-5)6x^3-x^2-13x-6(2x^2+3x+\frac{2}{3}\\ \underset{\_}{6x^3-10x^2}\\ \underset{\_}{\begin{array}{l}9x^2-13x-6\\ 9x^2-15x\end{array}}\\ 2x-6\\ \underset{\_}{2x-\frac{10}{3}}\\ \frac{10}{3}-6=\frac{-8}{3}\end{array}$

Thus, dividing $\left(-13x+6x^3-6-x^2\right)÷\left(3x-5\right)$ the quotient is $2x^2+3x+\frac{2}{3}$ and remainder is $\frac{-8}{3}$ .