Chapter 8.5, Problem 35WE

To determine

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

Answer to Problem 35WE

−2 is three times the root of P ( x ) = 0

Explanation of Solution

Given information: the information provided in the question,

  $P\left(x\right)=x^4+4x^2-16x-16;r=-2$

Calculation:

Divide P ( x ) by the factor obtained from r = −2, that is, $x+2$ if the remainder is 0 then it is a root at least once,

  $x+2\begin{array}{c}\hfill x^3+2x^2-4x-8\\ \hfill \overline{)\begin{array}{l}{x}^4+4x^3-16x-16\\ \underset{\_}{-x^4+2x^3}\end{array}}\end{array}$

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

  $\begin{array}{l}-4x^2-16x\\ \underset{\_}{4x^2+8x}\end{array}$

  $\begin{array}{l}-8x-16\\ \underset{\_}{8x+16}\end{array}$

0

Divide the quotient with $x+2$ , if remainder is 0, it is root at least twice,

  $x+2\begin{array}{c}\hfill x^2-4\\ \hfill \overline{)\begin{array}{l}{x}^3+2x^2-4x-8\\ \underset{\_}{-x^3-2x}\end{array}}\end{array}$

  $\begin{array}{l}-4x-8\\ \underset{\_}{4x+8}\end{array}$

  $\underset{\_}{0}$

Again, divide the above quotient with $x+2$ , if the remainder is 0, it is a rot for at least thrice,

  $x+2\begin{array}{c}\hfill x-2\\ \hfill \overline{)\begin{array}{l}{x}^2+0x-4\\ \underset{\_}{-x^2-2x}\end{array}}\end{array}$

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

  $\underset{\_}{0}$

Hence, −2 is three times the root of P ( x ) = 0