Question

Divide:$\frac{3a^2-8a-3}{a^2-25}÷\frac{3a^2-14a-5}{a^2+10a+25}$.

Short answer

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

Step-by-step solution

Step 1. Given information.

The given expression is:

$\frac{3a^2-8a-3}{a^2-25}÷\frac{3a^2-14a-5}{a^2+10a+25}$

Step 2. Simplify the expression.

$\frac{3a^2-8a-3}{a^2-25}÷\frac{3a^2-14a-5}{a^2+10a+25}$

Rewrite the division as the product of the first rational expression and the reciprocal of the second.

$\frac{3a^2-8a-3}{a^2-25}·\frac{a^2+10a+25}{3a^2-14a-5}$

Factor the numerators and denominators and then multiply.

$$\begin{gathered} \frac{\left(3a+1\right)\left(a-3\right)}{\left(a+5\right)\left(a-5\right)}·\frac{\left(a+5\right)\left(a+5\right)}{\left(3a+1\right)\left(a-5\right)} \\ \frac{\left(3a+1\right)\left(a-3\right)\left(a+5\right)\left(a+5\right)}{\left(a+5\right)\left(a-5\right)\left(3a+1\right)\left(a-5\right)} \end{gathered}$$

Simplify by dividing out common factors.

$\frac{\left(a-3\right)\left(a+5\right)}{\left(a-5\right)\left(a-5\right)}$