Question

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure to type the term with the variable first.The numerator is AnswerThe denominator is Answer

Answer 1

Explanation

$\begin{gathered} \frac{\left(q^2-9\right)}{\left(q^2+6q+9\right)}\div\frac{\left(q^2-2q-3\right)}{\left(q^2+2q-3\right)} \\ \end{gathered}$

Step 1

factorize:

remember those cases:

$\begin{gathered} (a^2-b^2)=(a+b)(a-b) \\ (a^2+2ab+b^2)=(a+b)^2 \\ \end{gathered}$

so

$\begin{gathered} (q^2-9) \\ (q^2-9)=(q^2-3^2)=(q+3)(q-3) \\ \text{and} \\ (q^2+6q+9)=(q^2+(2\cdot q\cdot3)+3^2)=(q+3)^2 \\ \text{also} \\ (q^2-2q-3)=(q+1)(q-3),\text{ because 1-3= -2 and, 1}\cdot-3=-3 \\ so \\ (q^2-2q-3)=(q+1)(q-3) \\ \text{ finally } \\ (q^2+2q-3)=(q-1)(q+3),because\text{ -1+3=}2,\text{ and -1}\cdot3=-3 \\ (q^2+2q-3)=(q-1)(q+3) \end{gathered}$

Step 2

replace

$\begin{gathered} \frac{(q^2-9)}{(q^2+6q+9)}\div\frac{(q^2-2q-3)}{(q^2+2q-3)} \\ \frac{(q+3)(q-3)}{(q+3)^2}\div\frac{(q+1)(q-3)}{(q-1)(q+3)} \\ \text{ reduce} \\ \frac{(q-3)}{(q+3)^{}}\div\frac{(q+1)(q-3)}{(q-1)(q+3)} \\ \frac{\frac{(q-3)}{(q+3)^{}}}{\frac{(q+1)(q-3)}{(q-1)(q+3)}}=\frac{(q-3)(q-1)(q+3)}{(q+3)(q+1)(q-3)} \\ \frac{(q-3)(q-1)(q+3)}{(q+3)(q+1)(q-3)}=\frac{(q-1)}{(q+1)} \\ \frac{(q-1)}{(q+1)} \end{gathered}$

therefore the answer is

numerator : q-1

denominator: q+1

I hope this helps you