Question

Divide Rational Expressions:$\frac{2n^2\hspace{0.17em}-\hspace{0.17em}3n\hspace{0.17em}-\hspace{0.17em}14}{25\hspace{0.17em}-\hspace{0.17em}{n}^2}÷\frac{2n^2\hspace{0.17em}-\hspace{0.17em}13n\hspace{0.17em}+\hspace{0.17em}21}{n^2\hspace{0.17em}-\hspace{0.17em}10n\hspace{0.17em}+\hspace{0.17em}25}$

Short answer

The required answer is$-\frac{\left(n+2\right)\left(n-5\right)}{\left(5+n\right)\left(n-3\right)}$.

Step-by-step solution

Step 1. Given information. 

The given expression is$\frac{2n^2\hspace{0.17em}-\hspace{0.17em}3n\hspace{0.17em}-\hspace{0.17em}14}{25\hspace{0.17em}-\hspace{0.17em}{n}^2}÷\frac{2n^2\hspace{0.17em}-\hspace{0.17em}13n\hspace{0.17em}+\hspace{0.17em}21}{n^2\hspace{0.17em}-\hspace{0.17em}10n\hspace{0.17em}+\hspace{0.17em}25}$.

Step 2.  Simplify the expression. 

$$\begin{gathered} \frac{2n^2\hspace{0.17em}-\hspace{0.17em}3n\hspace{0.17em}-\hspace{0.17em}14}{25\hspace{0.17em}-\hspace{0.17em}{n}^2}÷\frac{2n^2\hspace{0.17em}-\hspace{0.17em}13n\hspace{0.17em}+\hspace{0.17em}21}{n^2\hspace{0.17em}-\hspace{0.17em}10n\hspace{0.17em}+\hspace{0.17em}25} \\ =\frac{\frac{2n^2-3n-14}{25-n^2}}{\frac{2n^2-13n+21}{n^2-10n+25}} \end{gathered}$$

Apply the fraction rule.

$$\begin{gathered} \frac{\left(2n^2-3n-14\right)\left(n^2-10n+25\right)}{\left(25-n^2\right)\left(2n^2-13n+21\right)} \\ =\frac{\left(n+2\right)\left(2n-7\right)\left(n^2-10n+25\right)}{\left(25-n^2\right)\left(n-3\right)\left(2n-7\right)} \\ =\frac{\left(n+2\right)\left(n^2-10n+25\right)}{\left(25-n^2\right)\left(n-3\right)} \\ =\frac{\left(n+2\right){\left(n-5\right)}^2}{\left(25-n^2\right)\left(n-3\right)} \\ =\frac{\left(n+2\right){\left(n-5\right)}^2}{\left(5+n\right)\left(5-n\right)\left(n-3\right)} \\ =\frac{\left(n+2\right){\left(n-5\right)}^2}{-\left(5+n\right)\left(n-5\right)\left(n-3\right)} \\ =\frac{\left(n+2\right)\left(n-5\right)\left(n-5\right)}{-\left(5+n\right)\left(n-5\right)\left(n-3\right)} \\ =\frac{\left(n+2\right)\left(n-5\right)}{-\left(5+n\right)\left(n-3\right)} \\ =-\frac{\left(n+2\right)\left(n-5\right)}{\left(5+n\right)\left(n-3\right)} \end{gathered}$$