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

2 Answers
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

5
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#.

Was this helpful? Let the contributor know!
1500
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
NagrajB Share
Dec 16, 2015

Answer:

x cannot be zero
x cannot be 4

Explanation:

#(x)*(x^2 - 2x - 8)/(x(x-4))#
x can be cancelled in both numerator and denominator if and only if x not equal to zero

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

x - 4 cannot be cancelled if x = 0

Was this helpful? Let the contributor know!
1500