Chapter 11.3, Problem 21PPE

To determine

To divide $\left(2w^2+3w-15\right)÷\left(w-1\right)$ .

Answer to Problem 21PPE

Dividing $\left(2w^2+3w-15\right)÷\left(w-1\right)$ the quotient is $2w+5$ and remainder is $-10$ .

Explanation of Solution

Given:

  $\left(2w^2+3w-15\right)÷\left(w-1\right)$

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 $\left(2w^2+3w-15\right)÷\left(w-1\right)$ use long division method.

The division may be proceeded as:

  $\begin{array}{l}w-1)2w^2+3w-15(2w+5\\ \underset{\_}{2w^2-2w}\\ 5w-15\\ \underset{\_}{5w-5}\\ \underset{\_}{-10}\end{array}$

Thus, dividing $\left(2w^2+3w-15\right)÷\left(w-1\right)$ the quotient is $2w+5$ and remainder is $-10$ .