(x^2-81)/(x^2-16) divide (x+9)/(x-4)?

First, dividing by fractions is the same as multiplying by the reciprocal, so let's flip that around:

[(x² - 81) / (x² - 16)] * [(x - 4)/(x + 9)]

Now, both halves of the first fraction are both the difference of two squares, so let's extract those factors:

[(x + 9)(x - 9) / (x + 4)(x - 4)] * [(x - 4)/(x + 9)]

Now the x-4 can cancel out and the x+9 can, leaving:

(x - 9) / (x + 4)

(x^2 - 81)/(x^2 - 16) ÷ (x + 9)/(x - 4)
= (x + 9)(x - 9)/(x + 4)(x - 4) * (x - 4)/(x + 9) (canel out x - 4 , x + 9)
= (x - 9)/(x + 4)

I think you meant to write
(x^2-81)/(x^2-16) divided by (x+9)/(x-4)?

Fist we factor the two quadratic binomials

(x-9)(x+9)(x-4)(x+4) / (x+9)(x-4)

canceling out like terms in numerator and denominator


(x-9)(x+4) = x^2 -5x - 36

both 81 and 16 are squares so you can make that into 2.
(x+9)(x-9) and (x+4)(x-4)
Then, cancel out the bottom and you're left with (x-9)/(x+4)

(x + 9)(x -9)/(x +4)(x - 4) times (x -4)/(x + 9)

the (x -4)'s and the (x + 9)'s will cancel

(x -9)/(x +4) is the answer

a million