Chapter 8.5, Problem 36WE

To determine

To find: the number of times r is a root of P ( x ) = 0

Answer to Problem 36WE

−1 is four times is the root of P ( x ) = 0

Explanation of Solution

Given information: the information provided in the question,

  $P\left(x\right)=x^5+3x^4+2x^3-2x^2-3x-1;r=-1$

Calculation:

Divide P ( x ) by the factor obtained from r = −1, that is, $x+1$ if the remainder is 0 then it is a root at least once,

  $x+1\begin{array}{c}\hfill x^4+2x^3-2x-1\\ \hfill \overline{)\begin{array}{l}{x}^5+3x^4+2x^3-2x^2-3x-1\\ \underset{\_}{-x^5-x^4}\end{array}}\end{array}$

  $\begin{array}{l}2x^4+2x^3\\ \underset{\_}{-2x^4-2x^3}\end{array}$

  $\begin{array}{l}-2x^2-3x\\ \underset{\_}{2x^2+2x}\end{array}$

  $\begin{array}{l}-x-1\\ \underset{\_}{x+1}\end{array}$

0

Divide the quotient with $x+1$ , if remainder is 0, it is root at least twice,

  $x+1\begin{array}{c}\hfill x^3+x^2-x-1\\ \hfill \overline{)\begin{array}{l}{x}^4+2x^3-2x-1\\ \underset{\_}{-x^4-x^3}\end{array}}\end{array}$

  $\begin{array}{l}{x}^3\\ \underset{\_}{-x^3-x^2}\end{array}$

  $\underset{\_}{\begin{array}{l}-x^2-2x\\ x^2+x\end{array}}$

  $\underset{\_}{\begin{array}{l}-x-1\\ x+1\end{array}}$

Again, divide the above quotient with $x+1$ , if the remainder is 0, it is a root for at least thrice,

  $x+2\begin{array}{c}\hfill x^2-1\\ \hfill \overline{)\begin{array}{l}{x}^3+x^2-x-1\\ \underset{\_}{-x^3-x^2}\end{array}}\end{array}$

  $\begin{array}{l}-x-1\\ \underset{\_}{x+1}\end{array}$

  $\underset{\_}{0}$

  $x^2-1=\left(x-1\right)\left(x+1\right)$

Hence, −1 is four times is the root of P ( x ) = 0