Chapter 8.4, Problem 20WE

To determine

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

Answer to Problem 20WE

  $\begin{array}{l}Q(x)=x^3+2\\ R=5\end{array}$

Explanation of Solution

Given information :

The given equation:

  $x^4-5x^3+2x-5=(x-5)Q(x)+R$

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:

  $x^4-5x^3+2x-5=(x-5)Q(x)+R$

The polynomial (dividend) is $x^4-5x^3+2x-5$

The binomial (divisor) is $x-5$ .

To find the quotient and the remainder, divide $x^4-5x^3+2x-5$ by $x-5$ using synthetic division.

Use $5$ for c .

  $\begin{array}{l}\overline{)5}\text{ 1 }-5\text{ 0 2 }-\text{5}\\ \underset{\_}{\text{ 5 0 0 10}}\\ \text{ 1 0 0 2 | }5\end{array}$

The quotient is $x^3+2$ , and the remainder is $5$

Therefore,

  $2x^3+5x^2+4=(x+3)(x^3+2)+5$

Comparing with the given equation,

  $\begin{array}{l}Q(x)=x^3+2\\ R=5\end{array}$