In my Spivak Calc book, I've twice noticed a referrence to an 'algebraic trick,' but I don't understand it.
(8x^2 + 6x + 4)/ x+1 = 8x - 2 + (6/x+1)
and
((1+x)/(1-x))(1/x) = 2/(1-x) + 1/x
How does this trick work?
it's called polynomial long division, which works pretty much like regular division.
$\displaystyle 8x^2 + 6x + 4 = 8x^2 + 8x - 2x - 2 + 6 = 8x(x+1) - 2(x+1) + 6$.
so:
$\displaystyle \frac{8x^2 + 6x + 4}{x+1} = \frac{8x(x+1)}{x+1} - \frac{2(x+1)}{x+1} + \frac{6}{x+1}$
$\displaystyle = 8x - 2 + \frac{6}{x+1}$
what we are doing is writing a polynomial f(x) in the form:
f(x) = q(x)g(x) + r(x)
in order to deduce that:
$\displaystyle \frac{f(x)}{g(x)} = \frac{q(x)g(x) + r(x)}{g(x)} = \frac{q(x)g(x)}{g(x)} + \frac{r(x)}{g(x)}$
$\displaystyle = q(x) + \frac{r(x)}{g(x)}$
this is entirely analogous to writing for integers a,b:
a = qb + r
where q is called the QUOTIENT, and r is called the REMAINDER.
thus a/b = q + r/b.
for example, if a = 37, and b = 7, we have:
37 = (5)(7) + 2, so that:
37/7 = 5 + 2/7.
the long division of polynomials is carried out just as you would for integers, except instead of the "10s place" you have the "x place", and instead of the "100s place" you have the "x^{2} place" and so on.
the second example is slightly different.
note that:
$\displaystyle \frac{2}{1-x} + \frac{1}{x} = \frac{2x}{(1-x)x} + \frac{(1-x)}{(1-x)x} = \frac{2x + 1 - x}{(1-x)x} = \frac{1+x}{(1-x)x}$
$\displaystyle = \left(\frac{1+x}{1-x}\right)\left(\frac{1}{x}\right)$
(this is an example of "partial fraction decomposition" where one takes:
$\displaystyle \frac{f(x)}{g(x)h(x)}$
and splits it into the form:
$\displaystyle \frac{A(x)}{g(x)} + \frac{B(x)}{h(x)}$
in this case, the A(x) and B(x) turn out to be simple: A = 2, and B = 1, but this is not always the case (there may be more about this later on in the text when it comes time to integrate rational functions).).