Chapter 8.4, Problem 21WE

To determine

To find : the polynomial Q ( x ) and the constant R .

Answer to Problem 21WE

  $\begin{array}{l}Q(x)=2z^2-(3+4i)z+6i\\ R=2\end{array}$

Explanation of Solution

Given information :

The given equation:

  $2z^3-3z^2+8z-10=(z+2i)Q(x)+R$

Formula used :

In the division of a polynomial by a binomial

Dividend = (divisor) (quotient) $+$ remainder

The divisor in a synthetic division is a binomial of the form $x-c$ .

Calculation :

As per the problem,

In the given equation:

  $2z^3-3z^2+8z-10=(z+2i)Q(x)+R$

The polynomial (dividend) is $2z^3-3z^2+8z-10$

The binomial (divisor) is $z+2i$ .

To find the quotient and the remainder, divide $2z^3-3z^2+8z-10$ by $z+2i$ using synthetic division.

Use $-2i$ for c .

  $\begin{array}{l}\overline{)-2i}\text{ 2 }-3\text{ 8 }-10\\ \underset{\_}{-4i-8+6i\text{ 12}}\\ \text{ 2 }-3-4i6i\text{ | 2}\end{array}$

The quotient is $2z^2-(3+4i)z+6i$ , and the remainder is $2$

Therefore,

  $2z^3-3z^2+8z-10=(z+2i)(2z^2-(3+4i)z+6i)+2$

Comparing with the given equation,

  $\begin{array}{l}Q(x)=2z^2-(3+4i)z+6i\\ R=2\end{array}$