
Multiplication of Multinomials
The basis for multiplying two factors such as:
(x + 5)(2x − 3)
is the distributive rule:
a•(b + c) = a•b + b•c
We simply need to apply it several times:
(x + 5)(2x − 3) = (x + 5)•2x − (x + 5)•3 = x•2x + 5•2x − x•3 − 5•3
now collecting and combining like terms:
(x + 5)(2x − 3) = 2x^{2} + 7x − 15
Because this is a common situation, there is a mnemonic rule; FOIL which stands for:
First, Outer, Inner, Last
and indicates the four products we need to compute:
First—product of first two terms  x•2x = 2x^{2} 
Outer—product of two terms on the "outside"  x•(−3) = −3x 
Inner—product of two terms on the "inside"  5•2x = 10x 
Last—product of last two terms  5•(−3) = −15 
add, combine, and simplify  2x^{2} + 7x − 15 
Actually this is a special case of a general process to multiply two multinomials.
To multiply two multinomials multiply every term in the second factor by each term in the first factor. Add these. Then combine like terms and simplify. 
Let us suppose the first multinomial has m terms and the second has n terms. When we apply the rule, we will get a total of m•n terms. Of course, many of these may possibly be combined to get the final simplified expression.
For an example, consider:
(a + b + c)•(x + y)
Since the first factor has 3 terms and the second has 2 terms, we should expect to get a total of 6 terms:
multiplying the first term of the first factor by every term in the second  a•x + a•y 
multiplying the second term of the first factor by every term in the second  b•x + b•y 
multiplying the third term of the first factor by every term in the second  c•x + c•y 
adding them all together  a•x + a•y + b•x + b•y + c•x + c•y 
In this case, because we have used all different variables, the final expression cannot be simplified by combining terms.
As a more specific example, consider:
(x^{2} + 2x + 5)•(x + 3)
This problem also is to multiply a multinomial with 3 terms by a multinomial with 2 terms so we should expect to get 6 terms. Applying our process, we get:
multiplying the first term of the first factor by every term in the second  x^{2}•(x+3)  =  x^{2}•x + x^{2}•3  =  x^{3} + 3x^{2} 
multiplying the second term of the first factor by every term in the second  2x•(x+3)  =  2x•x + 2x•3  =  2x^{2} + 6x 
multiplying the third term of the first factor by every term in the second  5•(x+3)  =  5•x + 5•3  =  5x + 15 
adding them all together  x^{3} + 5x^{2} + 11x + 15 
For this example we initially get two terms containing an x^{2} and two terms with an x which can be combined to reduce the original six terms to a final four.
We have laid out the various products vertically; however, more typically the initial terms are written horizontally, then simplified:
(x^{2} + 2x + 5)•(x + 3) = x^{3} + 3x^{2} + 2x^{2} + 6x + 5x + 15 = x^{3} + 5x^{2} + 11x + 15
Either way is fine. The vertical layout is easier to visualize and to check intermediate results. The horizontal layout takes least room. Use whichever is easiest that can be done correctly!
Another alternative is arrange the terms in a table with those in the first factor down the rows and those in the second factor across the columns. Then fill in the table with the products:
 x  3 
x^{2}  x^{3}  3x^{2} 
2x^{ }  2x^{2}  6x^{ } 
5^{ }  5x^{ }  15^{ } 
and finally add all the individual pieces together and simplify:
x^{3} + 3x^{2} + 2x^{2} + 6x + 5x + 15 = x^{3} + 5x^{2} + 11x + 15
Whichever is the easiest method for you to perform the multiplication without error is the best for you!
The FOIL rule is simply a special case of this more general rule. In this case m = 2 and n = 2 so we should expect initially to get four terms—the four terms represented by the four letters of FOIL!
So how many terms should we initially get if we multiply the following?
(x^{4} + 2x^{3} − 5x^{2} + 3x + 4)(2x^{2} − x + 2)
Since we have five terms and three terms, we should get a total of 15 terms!
Because the highest power of x in the final result is x^{6} (which comes from multiplying the two respective terms with the highest powers in the two factors), and any sixth degree polynomial can be written with only seven terms (a term for each of the six different powers of x and a constant), we should expect to be able to combine some terms to reduce the 15 to 7 terms!
Try it and check your results!

