simplify (x^3-8)/(x^2-4)

Question

x³-8

x²-4

Answer 1

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

Can't divide by zero, so x^2-4 ≠ 0, x^2 ≠ 4, x ≠ ±2

x^3-8 = x^3 - 2^3 = (x-2)(x^2+2x+2^2)

x^2-4 = x^2 - 2^2 = (x+2)(x-2)

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

(x-2)/(x-2) = {UNDEFINED if x = 2, 1 if x ≠ 2}

There's a "Hole" at x = 2.

f(x) = (x^2+2x+4)/(x+2), x ≠ ±2

Divide by x+2:

x^2+2x+4 | x
x^2+2x
–––––––––
4

f(x) = x + 4/(x+2), x ≠ ±2

This is the simplest form for f(x) because it would be what you would use in integration in calculus.

It shows the Hole at (2,3), a vertical asymptote at x = -2, and a slant asymptote y = x.

check:

GeoGebra graph of both original and simplified functions:
http://www.wyzant.com/resources/files/266648/simplify_rational_function

Answer 2

 

 

(x^3-8) / (x^2-4)

 

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

 

what is left is x -2