Chapter 1.6, Problem 142E

To determine

To simplify: the rational expression $\frac{x^2+3x-40}{10+3x-x^2}$ .

Answer to Problem 142E

  $-\frac{\left(x+8\right)}{\left(x+2\right)}$ Where $x\ne -2$ and $x\ne 5$

Explanation of Solution

Given:

The rational expression: $\frac{x^2+3x-40}{10+3x-x^2}$

First factor the numerator and the denominator.

  $\begin{array}{c}\frac{x^2+3x-40}{10+3x-x^2}=\frac{x^2+8x-5x-40}{-x^2+5x-2x+10}\\ =\frac{x\left(x+8\right)-5\left(x+8\right)}{-x\left(x-5\right)-2\left(x-5\right)}\\ =\frac{\left(x+8\right)\left(x-5\right)}{\left(-x-2\right)\left(x-5\right)}\\ =-\frac{\left(x+8\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}\end{array}$

Now, find the values for which the denominator becomes 0, as division by 0 is undefined so for such values the given expression is not defined.

Now, the denominator $\left(x+2\right)\left(x-5\right)$ becomes 0 when either $x=-2$ or $x=5$ .

So, set the restriction on the domain of this expression to be $x\ne -2$ and $x\ne 5$ .

Lastly cancel out the common factor.

  $\begin{array}{c}\frac{x^2+3x-40}{10+3x-x^2}=-\frac{\left(x+8\right)\cancel{\left(x-5\right)}}{\left(x+2\right)\cancel{\left(x-5\right)}}\\ =-\frac{\left(x+8\right)}{\left(x+2\right)}\end{array}$

Conclusion:

The simplified form of the given rational expression is:

  $-\frac{\left(x+8\right)}{\left(x+2\right)}$ Where $x\ne -2$ and $x\ne 5$