How do you simplify $\frac { x ^ { 2} - 2x - 15} { x ^ { 2} - 9} \cdot \frac { x + 3} { x - 5}$ ?

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Answer 1 · 2016-10-20T21:18:42

$(x+3)/(x-3)$

$\frac { x ^ { 2} - 2x - 15} { x ^ { 2} - 9} \cdot \frac { x + 3} { x - 5}$

Factor the fraction on the left

What are the factors of -15 that add up to -2

$(x-5)(x+3)$

Difference of perfect squares

$(x^2-9)=(x^2-3^2)=(x+3)(x-3)$

Rewrite the expression using the factors you just found

$((x-5)(x+3))/((x+3)(x-3))*(x+3)/(x-5)$

Now cross cancel

$(cancel(x-5)cancel(x+3))/(cancel(x+3)(x-3))*(x+3)/cancel(x-5)$

You are then left with

$(x+3)/(x-3)$

Answer 2 · 2016-10-20T21:21:03

The expression can be simplified to $(x + 3)/(x- 3)$ .

$=((x - 5)(x + 3))/((x + 3)(x- 3)) xx (x + 3)/(x - 5)$

$=(x + 3)/(x - 3)$

Hopefully this helps!

How do you simplify \frac { x ^ { 2} - 2x - 15} { x ^ { 2} - 9} \cdot \frac { x + 3} { x - 5}\frac { x ^ { 2} - 2x - 15} { x ^ { 2} - 9} \cdot \frac { x + 3} { x - 5}?