Chapter 11.3, Problem 45PPE

To determine

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

Answer to Problem 45PPE

Dividing $\left(z^4+z^2-2\right)÷\left(z+3\right)$ the quotient is $z^3-3z^2+10z-10$ and remainder is 29..

Explanation of Solution

Given:

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

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

First, note that the terms of the dividend and divisor are already in descending powers of the variable.

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

  $\left(z^4+0z^3+z^2+0z-2\right)÷\left(z+3\right)$

Thus, the given division may be proceeded as:

  $\begin{array}{l}z+3)z^4+0z^3+z^2+0z-2(z^3-3z^2+10z-10\\ \underset{\_}{z^4+3z^3}\\ \underset{\_}{\begin{array}{l}-3z^3+z^2+0z-2\\ -3z^3-9z^2\end{array}}\\ 10z^2+0z-2\\ \underset{\_}{10z^2+20z}\\ -20z-1\\ \underset{\_}{-20z-30}\\ 29\end{array}$

Thus, dividing $\left(z^4+z^2-2\right)÷\left(z+3\right)$ the quotient is $z^3-3z^2+10z-10$ and remainder is 29.