Chapter 4.7, Problem 42WE

To determine

To find the multiple roots.

Answer to Problem 42WE

The values of multiple rootsare $\text{1 and 2}$

Explanation of Solution

Given information:

  ${\left(x^2-3x+2\right)}^3=0$

Calculation:

The given equation already has 0 as one side,and the other side is already factored. Simply set each factored equal to 0 and solve.

Thedouble rootscan be calculated as

  $\begin{array}{l}{\left(x^2-3x+2\right)}^3=0\\ \left(x^2-3x+2\right)\left(x^2-3x+2\right)\left(x^2-3x+2\right)=0\\ \left[x^2-\left(2x+x\right)+2\right]\left[x^2-\left(2x+x\right)+2\right]\left[x^2-\left(2x+x\right)+2\right]=0\left[\text{write }3x=\left(2x+x\right)\right]\\ \left[x^2-2x-x+2\right]\left[x^2-2x-x+2\right]\left[x^2-2x-x+2\right]=0\\ \left[x\left(x-1\right)-2\left(x-1\right)\right]\left[x\left(x-1\right)-2\left(x-1\right)\right]\left[x\left(x-1\right)-2\left(x-1\right)\right]\text{=0 }\\ \left[\left(x-1\right)\left(x-2\right)\right]\left[\left(x-1\right)\left(x-2\right)\right]\left[\left(x-1\right)\left(x-2\right)\right]=0\end{array}$

Therefore,

  $\begin{array}{l}\left(x-1\right)\text{=0 or }\left(x-2\right)=0\left[\text{use the zero-product property}\right]\\ x=1\text{ or }x=2\end{array}$

Hence,

The values of multiple rootsare $\text{1 and 2}$