If you really know your difference of two squares then you might spot
a short cut here:1
x− 1+1
x+ 1−1
x+ 2=2 x
x^2 − 1−1
x+ 2=2 x(x+ 2 )−(x^2 − 1 )
(x^2 − 1 )(x+ 2 ),etc.2.2.9 Division and the remainder theorem
➤
39 77 ➤Consider the sum
x+ 1 +3
x− 1Putting this over a common denominator gives
x+ 1 +3
x− 1=(x+ 1 )(x− 1 )+ 3
x− 1=x^2 − 1 + 3
x− 1=x^2 + 2
x− 1So adding polynomials to fractions presents few problems
(
it is analogous to, say, 2+^12 =
5
2). But what if we want to go the other way – that is, dividex−1intox^2 +2 to reproduce
the original sum? This is calledlong division. There is a routine algorithmic (i.e. step
by step) procedure for such division, but usually one can get away with a simplified
procedure which relies on repeatedly pulling out from the numerator terms that contain
the denominator. Our numerical example illustrates this:
5
2=4 + 1
2=2 × 2 + 1
2=2 × 2
2+1
2= 2 +1
2This approach requires only that we are good at spotting factors. So for example:
x^2 + 2
x− 1=x^2 − 1 + 3
x− 1=(x− 1 )(x+ 1 )+ 3
x− 1=(x− 1 )(x+ 1 )
x− 1+3
x− 1=x+ 1 +3
x− 1This does require some algebraic intuition and clever grouping of terms, but can be very
quick.