How do you simplify and find the restrictions for #(x^3-2x^2-8x)/(x^2-4x)#?

2 Answers
Sep 17, 2016

Answer:

#(x^3-2x^2-8x)/(x^2-4x)=x+2#, but #x# cannot take values #x=0# and #x=4#

Explanation:

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

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

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

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

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

= #(cancel(x)cancel((x-4))(x+2))/(1cancel(x)cancel((x-4)))#

= #x+2#

but #x# cannot take values #x=0# and #x=4# as that makes the denominator #x^2-4x=0#.

Sep 17, 2016

Answer:

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

Explanation:

Given:#" "(x^3-2x^2-8x)/(x^2-4x)#

The equation becomes 'undefined' if the denominator is 0. You are 'not allowed' to divide by 0.

#color(blue)("To determine the excluded value")#

Set #0=x^2-4x" " ->" " 0=x(x-4)#

Thus the excluded values for x are: #x=0" and "x=4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Simplifying the expression")#

Factor out #x# for top and bottom (numerator and denominator)

#=>(cancel(x)^1(x^2-2x-8))/(cancel(x)^1(x-4))" "=" "(x^2-2x-8)/(x-4)#

Factorizing the top

#=>((x+2)(x-4))/(x-4) = (x+2)xx(x-4)/(x-4)#

But #(x-4)/(x-4)=1#

Thus

#" "color(blue)(bar(ul(|color(white)(2/2)(x^3-2x^2-8x)/(x^2-4x) -= (x+2)color(white)(2/2)|)))#