Chapter 11.3, Problem 46PPE

To determine

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

Answer to Problem 46PPE

Dividing $\left(-2z^3-z+z^2+1\right)÷\left(z+1\right)$ the quotient is $-2z^2+3z-4$ and remainder is 5.

Explanation of Solution

Given:

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

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

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

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

Now, it is not required to insert placeholders with zero coefficients for missing powers of the variable as there are no missing powers.

Thus, the given division may be proceeded as:

  $\begin{array}{l}z+1)-2z^3+z^2-z+1(-2z^2+3z-4\\ \underset{\_}{-2z^3-2z^2}\\ \underset{\_}{\begin{array}{l}\text{ 3}{z}^2-z+1\\ 3z^2+3z\end{array}}\\ -4z+1\\ \underset{\_}{-4z-4}\\ 5\end{array}$

Thus, dividing $\left(-2z^3-z+z^2+1\right)÷\left(z+1\right)$ the quotient is $-2z^2+3z-4$ and remainder is 5.