Chapter 11.3, Problem 32PPE

To determine

To divide $\left(6x^4+4x^3-x^2\right)÷\left(6+2x\right)$ .

Answer to Problem 32PPE

Dividing $\left(6x^4+4x^3-x^2\right)÷\left(6+2x\right)$ the quotient is $3x^3-7x^2+\frac{41}{2}x+\frac{123}{2}$ and remainder is 369.

Explanation of Solution

Given:

  $\left(6x^4+4x^3-x^2\right)÷\left(6+2x\right)$

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 $\left(6x^4+4x^3-x^2\right)÷\left(6+2x\right)$ use long division method.

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

  $\left(6x^4+4x^3-x^2\right)÷\left(2x+6\right)$

The division may be proceeded as:

  $\begin{array}{l}2x+6)6x^4+4x^3-x^2(3x^3-7x^2+\frac{41}{2}x+\frac{123}{2}\\ \underset{\_}{6x^4+18x^3}\\ \underset{\_}{\begin{array}{l}-14x^3-x^2\\ -14x^3-42x^2\end{array}}\\ 41x^2\\ \underset{\_}{41x^2+123x}\\ -123x\\ \underset{\_}{-123x-369}\\ 369\end{array}$

Thus, dividing $\left(6x^4+4x^3-x^2\right)÷\left(6+2x\right)$ the quotient is $3x^3-7x^2+\frac{41}{2}x+\frac{123}{2}$ and remainder is 369.