How do you find the restricted values of x or the rational expression #(x^3-2x^2-8x)/(x^2-4x)#?

1 Answer
Dec 16, 2015

Answer:

#x!=0, 4#

Explanation:

Start by simplifying the equation:

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

#=(x(x^2-2x-8))/(x(x-4))#

#=(x(x-4)(x+2))/(color(red)xcolor(blue)((x-4)))#

Recall that any fraction cannot have a denominator of #0#. To find the restrictions for #x#, set each polynomial or term in the denominator to cannot equal to #0#, and solve for #x#.

Finding the restrictions

#1. color(red)x!=0#

#2. color(blue)(x-4)!=0#
#color(white)(ixxxx)x!=4#

#:.#, the restrictions are #x# are #x!=0# and #x!=4#.