Chapter 7.9, Problem 6MRE

To determine

To calculate: The simplified form of the expression $\frac{x^2-9}{x^2+5x+6}$ and state restrictions on variable.

Answer to Problem 6MRE

The simplified value of the expression $\frac{x^2-9}{x^2+5x+6}$ is $\frac{x-3}{x+2}$ and restriction is $x\ne -3,x\ne -2$ .

Explanation of Solution

Given information:

The expression $\frac{x^2-9}{x^2+5x+6}$ .

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ .

Calculation:

Consider the expression $\frac{x^2-9}{x^2+5x+6}$ .

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ .

Factorize numerator using $X^2-Y^2=\left(X+Y\right)\left(X-Y\right)$ and denominator using middle-term splitting method as,

  $\begin{array}{c}\frac{x^2-9}{x^2+5x+6}=\frac{\left(x+3\right)\left(x-3\right)}{x^2+3x+2x+6}\\ =\frac{\left(x+3\right)\left(x-3\right)}{x\left(x+3\right)+2\left(x+3\right)}\\ =\frac{\left(x+3\right)\left(x-3\right)}{\left(x+3\right)\left(x+2\right)}\\ =\frac{x-3}{x+2}\end{array}$

To restrict the variable, exclude the value that makes the denominator zero.

The denominator will be zero when $\begin{array}{c}\left(x+3\right)\left(x+2\right)=0\\ x+3=0,x+2=0\\ x=-3,x=-2\end{array}$ .

For denominator not to be zero, $x\ne -3,x\ne -2$ .

Thus, the simplified value of the expression $\frac{x^2-9}{x^2+5x+6}$ is $\frac{x-3}{x+2}$ and restriction is

  $x\ne -3,x\ne -2$ .