simplifying rational expressions please help!

(n^2 + 8n + 15) / (n^2 - 25)

First, factor whatever you can.

n^2 + 8n + 15 = (n + 3)(n + 5)

n^2 - 25 = (n - 5)(n + 5)

So you have...

(n + 3)(n + 5) / (n - 5)(n + 5)

The (n + 5) on top and bottom will cancel...

= (n + 3) / (n - 5)

Factorise above eqn : n^2+8n+15 = > (n+5)(n+3)

Factorise below eqn : n^2-25 => (n+5)(n-5)

so : (n+3)/(n-5)

= (n² + 8n + 15)/(n² - 25)

= ([n + 5][n + 3])/([n + 5][n - 5])

= ([{n + 5}/{n + 5}][n + 3])/([{n + 5}/{n + 5}][n - 5])

= (n + 3)/(n - 5)

Answer: (n + 3)/(n - 5)

Checking:

= ([n + 3]/[n - 5])(n² - 25)

= ([n + 3]/[n - 5])([n - 5][n + 5])

= (n + 3)(n + 5)

= n² + 5n + 3n + 15

= n² + 8n + 15

(n^2 + 8n + 15)/(n^2 - 25)

= (n^2 + 5n + 3n + 15)/(n^2 + 5n - 5n - 25)

= (n + 5)(n + 3)/(n + 5)(n - 5) (cancel out n + 5)

= (n + 3)/(n - 5)

for x^2-12x+35 find 2 numbers which the multiply is 35 and sum is -12 so we have -7 & -5 for x^2-3x-28 ind 2 numbers which the multply is -28 and sum is -3 so we have 4 & -7 (x-7)(x-5) -------------- (x-7)(x+4) so (x-7) will cancel the we will have (x-5) --------- (x-4)

(n+3)(n+5)

---------------

(n+5)(n-5)

n+3

------

n-5

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