# Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises - Page 53: 14

$\color{blue}{(-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, 5) \cup (5, +\infty)}$

#### Work Step by Step

The denominator of a rational expression is not allowed to be equal to zero as it will make the expression undefined. Find the values of $x$ for which the denominator is equal to zero by equating the denominator to zero: $(2x+3)(x-5)= 0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: \begin{array}{ccc} &2x+3 = 0 &\text{ or } &x-5=0 \\&2x=-3 &\text{or} &x=5 \\&x=-\frac{3}{2} &\text{or} &x=5 \end{array} This means that the value of $x$ can be any real number except $-\frac{3}{2}$ and $5$. Therefore, the domain of the given rational expression is the set of real numbers except $-\frac{3}{2}$ and $5$. In interval notation, the domain is: $\color{blue}{(-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, 5) \cup (5, +\infty)}$

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