Question 185333
<font face="Times New Roman" size="+2">


FOIL *[tex \Large \ \ \Rightarrow\ \] First, Outside, Inside, Last


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x + 1)(x - 2)]


The First terms are:  <i>x</i> and <i>x</i>.  *[tex \Large \ \ \ ] So: *[tex \Large x\ \cdot\ x = x^2]


The Outside terms are: <i>x</i> and -2. *[tex \Large \ \ \ ]So: *[tex \Large x\ \cdot\ -2 = -2x]


The Inside terms are: 1 and <i>x</i>. *[tex \Large \ \ \ ]So: *[tex \Large 1\ \cdot\ x = x]


The Last terms are: 1 and -2. *[tex \Large \ \ \ ]So: *[tex \Large 1\ \cdot\ -2 = -2]


Add all four results:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + (-2x) + x + (-2)]


Collect terms and remove parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - x - 2]


Note that there is nothing sacred about the order in which you do the steps; you would still end up with the correct answer if you did Outside, Inside, Last, First.  It is just that OILF doesn't spell anything and is therefore harder to remember.  Also, doing things in that order, given that your binomials are arranged in decending powers of the variable, results in the product trinomial arranged in decending powers of the variable, i.e. standard form.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>