# Different way to do Algebra

1. Jun 23, 2012

### livindesert

Is their a different way to solve basic polynomials without the F.O.I.L. method?

2. Jun 23, 2012

### Muphrid

You mean to multiply polynomials? Well, the order of the terms isn't important, but you need all the relevant terms.

This is like asking if there's a different way to mutiply 27 x 68. In the end, you're sill going to get four terms: 1200 + 160 + 420 + 56. That we're generally taught to start with the last (56) and to carry (adding incrementally instead of getting four separate terms) doesn't change the fundamental nature of the process.

3. Jun 23, 2012

### Ratch

livindesert,

What does F.O.I.L. mean?

Ratch

4. Jun 23, 2012

### Number Nine

"First, outer, inner, last", I assume, referring to the process of multiplying two binomials.

Fun(ny) fact: One of my abstract algebra professors (discussing polynomials as a unique factorization domain) once threatened physical harm on a student who mentioned "FOIL" in class.

5. Jun 24, 2012

### HallsofIvy

The general rule is, when you are multiplying sums, each term of the first sum must multiply each term of the second and then added.

In particular, if both sums have two members, (ax+ b)(cx+ d), The "ax" must multiply both "cx" and "d" giving $acx^2$ and $adx$. The "b" must multiply both "cx" and "d" giving $bcx$ and $bd$. Adding those, $(ax+ b)(cx+ d)= acx^2+ adx+ bcx+ bd= acx^2+ (ad+ bc)x+ bd$.

"FOIL" just makes sure all of those four multiplications are done. If we have two sums each with three terms, say (ax+ by+ c)(dx+ ey+ f), "each term in the first sum multiplies each term in the second sum", gives $adx^2+ aexy+ afx+ bdxy+ bey^2+ bfy+ cdx+ cdy+ cf= adx^2+ (ae+ bd)xy+ bey^2+ (af+ cd)x+ (bf+ cd)y+ cf$

6. Jun 24, 2012

### Millennial

In general, multiplying polynomials follow directly from the distributive property of multiplication, that is, $a(b+c)=ab+ac$. This follows from the definition of multiplication. Now, we apply it to an arbitrary product of polynomials as follows. Let's say we have $(2a+b)(3c+2d)$. What we do is to apply the rule once to get $2a(3c+2d)+b(3c+2d)$ and then apply it once more on each of the terms to get $6ac+4ad+3bc+2bd$. This concept can be generalized to more terms in polynomials and more polynomials as well.