I've multiplied it out so that they both have a common denominator of (x+4)(x-2) and now I'm confused on what to do next.

Simplify: (2x - 5)/(x - 2) + (x - 8)/(x + 4) = f(x)

First, because you can't divide by zero,

note that x ≠ -4 or 2.

You have two fractions.

To add them you need a common denominator.

f(x) = (2x - 5)(x + 4)/((x + 4)(x - 2))

+ (x - 8)(x - 2)/((x - 2)(x + 4))

f(x) = ((2x - 5)(x + 4) + (x - 8)(x - 2))/((x - 2)(x + 4))

f(x) = (2x(x + 4) - 5(x + 4) + x(x - 2) - 8(x - 2))

/(x(x + 4) - 2(x + 4))

f(x) = (2x^2 + 8x - 5x - 20 + x^2 - 2x - 8x + 16)

/(x^2 + 4x - 2x - 8))

f(x) = (3x^2 - 7x - 4)/(x^2 + 2x - 8))

First, because you can't divide by zero,

note that x ≠ -4 or 2.

You have two fractions.

To add them you need a common denominator.

f(x) = (2x - 5)(x + 4)/((x + 4)(x - 2))

+ (x - 8)(x - 2)/((x - 2)(x + 4))

f(x) = ((2x - 5)(x + 4) + (x - 8)(x - 2))/((x - 2)(x + 4))

f(x) = (2x(x + 4) - 5(x + 4) + x(x - 2) - 8(x - 2))

/(x(x + 4) - 2(x + 4))

f(x) = (2x^2 + 8x - 5x - 20 + x^2 - 2x - 8x + 16)

/(x^2 + 4x - 2x - 8))

f(x) = (3x^2 - 7x - 4)/(x^2 + 2x - 8))

Do long division:

3x^2 - 7x - 4 | 3

3x^2 + 6x -24

-13x + 20 | 0

f(x) = 3 - (13x - 20)/((x - 2)(x + 4))

Find partial fractions:

(13x - 20)/((x - 2)(x + 4)) = A/(x - 2) + B/(x + 4)

13x - 20 = A(x + 4) + B(x - 2) = (A + B)x + 4A - 2B

A + B = 13

4A - 2B = -20

2A + 2B = 26

6A = 6

A = 1

B = 12

f(x) = 3 - (1/(x - 2) + 12/(x + 4))

f(x) = 3 - 1/(x - 2) - 12/(x + 4)

This is the simplest form of f(x) and is the one to use if you want to integrate f(x). Calculus is the goal; right?!

check:

Graph the initial form of f(x) and the last form to see if they are the same. See: http://www.wyzant.com/resources/files/265277/simplify_to_partial_fractions √