Chapter 4.6, Problem 41WE

To determine

To find: The solution set of the equation $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ .

Answer to Problem 41WE

The solution set for the equation $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ is $\left\{6\right\}$ .

Explanation of Solution

Given information:

The equation is $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ .

Calculation:

The given equation is $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ .

Solve the given equation $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ to obtain the value of solution set.

  $\begin{array}{c}\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)\\ x^3-2x^2+6x-3x^2+6x-18=x^3-5x^2+9x\\ x^3-5x^2+12x-18=x^3-5x^2+9x\left(\text{merge the like terms}\right)\\ 12x-18=9x\left(\text{subtract }{x}^3\text{ and }-5x^2\text{ from both sides}\right)\end{array}$

Further solve the above calculation,

  $\begin{array}{c}3x-18=0\left(\text{subtract 9}x\text{ from both sides}\right)\\ \text{3}x=18\left(\text{add 18 on both sides}\right)\\ x=\text{6 }\left(\text{divide both side by 3}\right)\end{array}$

Therefore, the solution set for the equation $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ is $\left\{6\right\}$ .

Conclusion:

Thus, the solution set for the equation $\left(x-3\right)\left(x^2-2x+6\right)=\left(x\right)\left(x^2-5x+9\right)$ is $\left\{6\right\}$ .