8.2 Multiply and divide rational expressions  (Page 2/3)

Solution

Solution

Rewrite the division as the product of the first rational expression and the reciprocal of the second.
Factor the numerators and denominators and then multiply.
Simplify by dividing out common factors.

Solution

$\begin{array}{cccc}& & & \hfill \frac{2x^2+5x-12}{x^2-16}÷\frac{2x^2-13x+15}{x^2-8x+16}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite the division as multiplication of}\hfill \\ \text{the first expression by the reciprocal of}\hfill \\ \text{the second.}\hfill \end{array}\hfill & & & \hfill \frac{2x^2+5x-12}{x^2-16}·\frac{x^2-8x+16}{2x^2-13x+15}\hfill \\ \\ \\ \begin{array}{c}\text{Factor the numerators and denominators}\hfill \\ \text{and then multiply.}\hfill \end{array}\hfill & & & \hfill \frac{\left(2x-3\right)\left(x+4\right)\left(x-4\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)\left(2x-3\right)\left(x-5\right)}\hfill \\ \\ \\ \text{Simplify by dividing out common factors.}\hfill & & & \hfill \frac{\cancel{\left(2x-3\right)}\cancel{\left(x+4\right)}\cancel{\left(x-4\right)}\left(x-4\right)}{\cancel{\left(x-4\right)}\cancel{\left(x+4\right)}\cancel{\left(2x-3\right)}\left(x-5\right)}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \hfill \frac{x-4}{x-5}\hfill \end{array}$

Solution

$\begin{array}{cccc}& & & \hfill \frac{p^3+q^3}{2p^2+2pq+2q^2}÷\frac{p^2-q^2}{6}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite the division as a multiplication}\hfill \\ \text{of the first expression times the}\hfill \\ \text{reciprocal of the second.}\hfill \end{array}\hfill & & & \hfill \frac{p^3+q^3}{2p^2+2pq+2q^2}·\frac{6}{p^2-q^2}\hfill \\ \\ \\ \begin{array}{c}\text{Factor the numerators and denominators}\hfill \\ \text{and then multiply.}\hfill \end{array}\hfill & & & \hfill \frac{\left(p+q\right)\left(p^2-pq+q^2\right)6}{2\left(p^2+pq+q^2\right)\left(p-q\right)\left(p+q\right)}\hfill \\ \\ \\ \text{Simplify by dividing out common factors.}\hfill & & & \hfill \frac{\cancel{\left(p+q\right)}\left(p^2-pq+q^2\right){\cancel{6}}^3}{\cancel{2}\left(p^2+pq+q^2\right)\left(p-q\right)\cancel{\left(p+q\right)}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \hfill \frac{3\left(p^2-pq+q^2\right)}{\left(p-q\right)\left(p^2+pq+q^2\right)}\hfill \end{array}$

Solution

$\begin{array}{cccc}& & & \hfill \frac{a^2-b^2}{3ab}÷\left(a^2+2ab+b^2\right)\hfill \\ \\ \\ \text{Write the second expression as a fraction.}\hfill & & & \hfill \frac{a^2-b^2}{3ab}÷\frac{a^2+2ab+b^2}{1}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite the division as the first}\hfill \\ \text{expression times the reciprocal of the}\hfill \\ \text{second expression.}\hfill \end{array}\hfill & & & \hfill \frac{a^2-b^2}{3ab}·\frac{1}{a^2+2ab+b^2}\hfill \\ \\ \\ \begin{array}{c}\text{Factor the numerators and the}\hfill \\ \text{denominators, and then multiply.}\hfill \end{array}\hfill & & & \hfill \frac{\left(a-b\right)\left(a+b\right)·1}{3ab·\left(a+b\right)\left(a+b\right)}\hfill \\ \\ \\ \text{Simplify by dividing out common factors.}\hfill & & & \hfill \frac{\left(a-b\right)\cancel{\left(a+b\right)}}{3ab·\cancel{\left(a+b\right)}\left(a+b\right)}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \hfill \frac{\left(a-b\right)}{3ab\left(a+b\right)}\hfill \end{array}$