Question

How do you multiply and simplify `\frac { 2a } { a ^ { 2} - 81} \cdot \frac { a - 9} { a }`?

Answer 1

`\frac { 2} { a+9)`

Explanation:

`\frac { 2a } { a ^ { 2} - 81} \cdot \frac { a - 9} { a }`

Instead of just multiplying across like we would with any pair of fractions, let's realize that `a^2-81` can be factored to simplify our multiplication.

Let's rewrite our problem like this...

`\frac { (2a)(a-9) } { (a ^ { 2} - 81)(a)`

Divide common terms `2a` and `a`

`\frac { 2cancela(a-9) } { (a ^ { 2} - 81)cancel(a)`

`(2(a-9) }/ { (a ^ 2 - 81)`

Now factor `a^2-81` as a `"Difference of perfect squares"`
[Here is an awesome video if you are confused on how to factor difference of perfect squares.](https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-polynomials-3-special-product-forms/v/factoring-difference-of-squares)

`\frac { 2(a-9) } { (a+9)(a-9)`

Cancel out equal terms

`\frac { 2cancel((a-9)) } { (a+9)cancel((a-9))`

and our final answer is

`\frac { 2} { a+9)`