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

Question

Avery Alexander · Accepted Answer

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

Avery Alexander · Answer

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

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)|)))#

Abigail Allen · Answer

undefined To simplify the expression (x^3-2x^2-8x)/(x^2-4x), we can factor out an x from the numerator and denominator, giving us x(x^2-2x-8)/(x(x-4)).

Next, we can simplify the expression in the numerator by factoring the quadratic x^2-2x-8. This factors as (x-4)(x+2).

Substituting this back into the expression, we have x(x-4)(x+2)/(x(x-4)).

Now, we can cancel out the common factors of (x-4) and x in the numerator and denominator, leaving us with (x+2)/x as the simplified expression.

To find the restrictions, we need to identify any values of x that would make the denominator equal to zero. In this case, x cannot be equal to zero or 4, as these values would result in division by zero. Therefore, the restrictions for this expression are x ≠ 0 and x ≠ 4.