Question #10b08

1 Answer
Apr 4, 2017

Your answer is incorrect. A correct simplification could include #-(3x+2)/(x^2-4)#.

Explanation:

As with most fractions, we should try to find a common denominator.

The first fraction's denominator can be factored as a difference of squares:

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

We now should see that the least common denominator is #(x+2)(x-2)#. So, we need to multiply the second fraction by #(x+2)# in the numerator and denominator:

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

Simplify the numerator of the second fraction by FOILing #(x+1)(x+2)#:

#x^2/((x+2)(x-2))-((x+1)(x+2))/((x+2)(x-2))=x^2/((x+2)(x-2))-(x^2+3x+2)/((x+2)(x-2))#

Since the fractions have the same denominator, we can combine the numerators. Be careful, though--since the second fraction is being subtracted, we will have to use parentheses and subtract the entire numerator of the second denominator:

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

Distributing the negative:

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

Canceling:

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

This could also be written as any of the following:

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