Simplify the following rational expression. List all of the excluded values, and classify each as a removable discontinuity, or a nonremovable discontinuity. -2/x^2-4+x-1/x^2-2x

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Excluded values  -2,0,2  These are non removable since we cannot cancel them in the numerator

(x+1)

x(x+2)  

Step-by-step explanation:

-2                  x-1

+

x^2 -4        x^2 -2x

First factor the denominators

-2                  x-1

+

x^2 -4        x^2 -2x

x^2 -4 is the difference of squares

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

x^2 -2x has an x in common

x(x-2)

-2                   x-1

+    

(x-2)(x+2)        x(x -2)

The values that are excluded are the values that make the denominators zero

(x-2) (x+2) = 0     x=2  x=-2

x(x-2) =0  x=0  x=2

Excluded values  -2,0,2  These are non removable since we cannot cancel them in the numerator

We need to get a common denominator of (x-2)(x+2)x

Multiply the first term by x/x and the second term by (x+2)/(x+2)

-2 x                 (x-1) (x+2)

+    

x(x-2)(x+2)        x(x -2)(x+2)

Distribute

(x+2)(x-1) = x^2 -x+2x-2 = x^2+x-2

-2 x                x^2+x-2

+    

x(x-2)(x+2)        x(x -2)(x+2)

Combine like terms, since the denominators are the same

-2 x  + x^2+x-2

x(x-2)(x+2)    

 x^2-x-2

x(x-2)(x+2)      

Factor the numerator

 x^2-x-2 = (x-2) (x+1)

(x-2) (x+1)

x(x-2)(x+2)  

Cancel like terms

(x+1)

x(x+2)  

The excluded values are found at the beginning, not now.  We canceled some of them in our simplified expression.  They are still excluded values

no, because the sides are not congruent