Reducing fractions

In the numerator, an expression can be factored out. To see that, it helps to reorder the terms in the more common order (starting with the highest power of x). The signs have to be considered carefully.

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

Therefore, we may continue to simplify: 
 

\(\frac{(-x+1)(x+2)}{x-1} = \frac{(-1)(x-1)(x+2)}{x-1} = (-1)(x+2) = -x-2\)

\(\frac{x^2-b^2}{x+b} = \frac{(x-b)(x+b)}{x+b} = x-b\)

Yes! The numerator is a binomial formula of the third kind. Therefore, it can be factorized and thereafter the fraction can be reduced. 

\(\frac{x^2+b^2}{x+b}\) cannot be reduced! The numerator is no binomial formula, and nothing can be factored out. Therefore, this fraction of sums cannot be reduced. 
 

Right!

There is a sum in the denominator. Let us assume y=3, for example (as y represents a number that we can choose). Then the fraction becomes \(\frac{4}{4+3}\) 

Here, it is obvious that the fraction cannot be reduced and the numbers in the denominator must simply be added up. 

This is the only thing we can do; however, we do not know the value of y, therefore we must leave the denominator 4+y as it stands.

Terribly wrong!

There is a sum in the numerator, from which no parts can be reduced. 

To show that, let us assume a=1 (a could be any number, so a could be 1). Then, our claim becomes: 

\(\frac{4+3}{4}=\frac{1+3}{1}=4\)

which is obviously false. 

There is no way around it: both numbers of the sum in the numerator must be divided by the denomiator.
 

No!

There is a sum in the numerator, from which nothing can be factored out.