## Precalculus (6th Edition) Blitzer

We have the given algebraic expression: $\frac{{{x}^{2}}-25}{x-5}=x-5$ We know that for an algebraic expression, a rational expression is an expression which can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$. By solving the left-hand side of the given expression, we have: $\frac{{{x}^{2}}-25}{x-5}$ And factorize ${{x}^{2}}-25$ using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. \begin{align} & {{x}^{2}}-25={{\left( x \right)}^{2}}-{{\left( 5 \right)}^{2}} \\ & =\left( x+5 \right)\left( x-5 \right) \end{align} So, the left-hand side of the given expression becomes: $\frac{{{x}^{2}}-25}{x-5}=\frac{\left( x+5 \right)\left( x-5 \right)}{\left( x-5 \right)}$ And solve further to obtain the solution: \begin{align} & \frac{{{x}^{2}}-25}{x-5}=\frac{\left( x+5 \right)\left( x-5 \right)}{\left( x-5 \right)} \\ & =x+5 \end{align} Hence, $\frac{{{x}^{2}}-25}{x-5}=$ $x+5$.