"FOIL" and factoring

FOIL is a shorthand quick trick for one specific special case of multiplication and factoring, and unless your class is really bad and only covers this one case I don’t recommend it, a dead end that is more often wrong than right.

Factoring is just **undoing multiplication**
You know that 3 x 5 = 15, so that when you are doing things like breaking down fractions you break up 15 into its *factors* 3 x 5
Same idea in algebra, just more terms. But you;re always just breaking a multiplication back apart.

In algebra, it is very important to understand the Distributive Law.
To distribute means to spread things around or to share things out.
The Distributive Law says that when you multipy some kind of grouped quantity in a parenthesis or other grouping symbol, you share the multiplication around *all* parts of it.
Actually we use this in arithmetic all the time too.
Look at the multiplications 2 x 14 and 32 x 13
You do this (hope the spacing prints OK)

14
2
__
28

13
32
__
26
39
____
416

So, what are you doing? In 2 x 14, you multiply the 2 x 4 in the units or ones and put the 8 there, then the 2 x 1 in the tens (actually 2 x 10 = 20 and use the place value to keep the tens straight) and put the 2 in the tens place. You have *distributed* the 2 over the 10 and the 4, multiplying each and adding the results together at the end.

In 32 x 13, first you multiply 2 x 3 in the units or ones and put down a 6, then 2 x 1 (actually 2 x 1 ten) in the tens and put 2 in the tens place. You have *distributed* the 2 x over the 1 ten plus 3.
Then you multiply the 3 (actually 3 tens or 30) times the 3 in the ones place and get 9, which you put in the tens place because it’s really 30 x 3 = 90; then you multiply the 3 tens times the 1 ten and put the results in the hundreds place because tens times tens = hundreds. You have *distributed* the 30 over the tens and ones.
Then at the end after multiplying you add it all up.

Our usual system does all the place value automatically, but this is what is going on.
Note that when you multiply two digits by two digits you get *four* sub-products that you have to add all together.

If we write this out algebra style, you would write 32 = 30 + 2 and 13 = 10 + 3
The 32 x 13 = (30 + 2) x (10 + 3) = 30x10 + 30x3 + 2x10 + 2x3
= 300 + 90 + 20 + 6 = 300 + 110 + 6 = 416, same as above, see it works!

The whole FOIL thing is a mnemonic to remember all four parts of this two-digit times two digit multiplication. Trouble is that sometimes you multiply one digit by two digits or one digit times three digits or two digits times three digits, and *in all of these cases* FOIL DOES NOT WORK, it is dead wrong. I disrecommend learning a “standard” technique that is wrong more often than it is right.

A much more productive rule is:
Multiply EVERYTHING in the first bracket by EVERYTHING in the second bracket; THEN ADD the products together.
As a doublecheck, the number of sub-priducts *before* you add should be the product of the number of terms in each bracket, for example two terms times three terms = six sub-products before adding and combining like terms.
Always check signs + or - as you multiply each product. Do signs first, then numbers (coefficients), then letters, for each sub-product.

Some examples, writing x2 to mean “x squared”, sorry the format is limited.

a(b + c) = ab + ac [one term times two terms, NOT a FOIL, won’t work]

3(x - y) = 3x - 3y [one term times two terms, NOT a FOIL, won’t work]

(a + b)(c + d) = ac + ad + bc + bd [Two terms times two terms; this is the ONE special case where FOIL makes sense.]

(x + 3)(x - 5) = x2 - 5x + 3x - 15 = x2 - 2x - 15, combining like terms [Two terms times two terms; this is the ONE special case where FOIL makes sense.]

5y(y2 - 3y -1) = 5y3 - 15y2 - 5y [one term times three terms, NOT a FOIL, won’t work]

(a - 1)(a2 - 4a + 6) = a3 - 4a2 + 6a - a2 + 4a - 6 = a3 - 5a2 + 10a - 6
combining like terms at end [two terms times three terms, NOT a FOIL, won’t work]

So why am I giving all these examples of multiplication when you asked about factoring? Because they are the same thing!!
Factoring is asking OK, if I split this into *multiplied* parts called factors, what was multiplied before I got here and came on the case?

Look at 7x2 - 14x
Hmm, says the detective, I see a factor of 7 in both terms. And there is an x in both terms. I can split out the 7x called a *common factor* and see what it was multiplied and distributed over
7x2 - 14x = 7x(x - 2)
If I multiplied that back, 7x(x - 2) = 7x2 - 14x back with the original, yup this works!

Look at x2 + 8x + 15
Hmm says the detective, there’s nothing common here, so can I split it up another way? Let’s think, the beginning x2 is of course x times x, that’s what x2 means. The 15 is 3 times 5; now can I make the middle work out?
Multiply and see: (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15 combining like terms, and hey, it works! (This one is a two digit by two digit, the ONLY case where FOIL makes sense.)
So we factor x2 + 8x + 15 = (x + 3)(x + 5)

More complicated problems like 4y2 - 23y - 6, well, you look at *all* the options and see what works. 4y2 could be 2y times 2y OR it could be 4y times y. 6 could be 2 times 3 OR it could be 6 times 1. You do know there’s one plus and one minus because that’s how you get the negative product -6. You try multiplying all the pairs possible and see what comes up.
(2y + 2)(2y - 3) = 4y2 -6y + 4y - 6 = 4y2 - 2y - 6 Nope, not quite right, that middle term is off.
(2y + 1)(2y - 6) = 4y2 -12y +2y - 6 = 4y2 - 10y - 6 Nope, not quite right, that middle term is off.
(4y + 2)(y - 3) = 4y2 - 12y + 2y - 6 = 4y2 -10y -6 Nope, not quite right, that middle term is off.
(4y + 3)(y - 2) = 4y2 - 8y + 3y - 6 = 4y2 - 5y - 6 Nope, not quite right, that middle term is off.
(4y + 6)(y - 1) = 4y2 -4y + 6y - 1= 4y2 + 2y -1 Nope, not quite right, that middle term is off.
(4y + 1)(y - 6) = 4y2 -24y + y - 6 = 4y2 - 23y - 6 GOT IT!!

so we write the factoring 4y2 - 23y - 6 = (4y + 1)(y - 6)

Note that if you get good at the multiplication, the factoring pops out at you; if you can’t do the multiplication, the factoring is very very unlikely to come out; so go back and get that multiplication thoroughly mastered.

Note also that we already figured out the first and the last terms by plan, it’s only the middle term that bothers us, so we can save a lot of wasted effort by just checking that middle term.

There are various tricks that people try, but they come out longer than just working it out, I promise.

This is the basics; other special cases when you’re ready.

I am at the “Uni” right now, when I get home, I am going to go over what all you wrote out, I printed it and everything. Thank you very much.

I had a student who taught me several nifty tricks he’d developed. Basically, when processes had a zillion steps and could get spatially confusing, he’d mark things as he went.
So for something like a(b+6) he’d draw an arrow from the a to the b and also from the a to the 6 to make sure he multiplied by both.
When he had to multiply or divide an entire side of an equation by something, he’d write the number (small :-)) next to each thing he had to multiply it by.

One thing that might help conceptually is to imagine that each number in the first group has to take a turn and “dance” (multiply) wiht each number in the next group.

We have a flowchart that takes you through the possible steps for factoring (starting with looking for a common factor in everything — so when you have 6x^3 + 3x^2 +9x, the first thing you would do would be to take ouit 3x) — Monday when I’m back at work I could scan it and post it online, if you’d like to see it. It’s one of those things that some students like & some students just find too complicated, but it includeds things like the common patterns that shoujld be committed to memory so that when you see “4x^2 - 9” you automatically say “Oh! Difference of Squares!”

Wish you were here!!!!! I’ve got half a dozen folks doing *exactly* this stuff…

Victoria and Sue I thank the both of you very much. And Sue, if it you do not mind, please scan that sheet you were speaking of. I took my final today and I really and truly think that I did all right. I had a professor who did not do partial credit, but my very last test I basically graded myself…I knew I missed six problems and I did my own score…normally I never get an answer that is like A B or C…it was funny because I really never know if I am right or wrong.

Victoria, I thank you for your post, and I am sorry for not reading Sue’s sooner…but I have had finals week, you know? So Sue, I am sorry for not having had the time to read your post sooner…Victoria, I printed out your post and I might just frame it or something:) I have it in page protectors in my big binder of mathematics, where I cross reference the topics and all of that…page protectors, my friend! I like that whole concept of the numbers dancing, I almost understand that.

I hate to do this, but…I saved my scratch paper from my final. Alright, there is this one problem that I just figured was prime, but the answer just was something I could not see, I think I could not really see how to do it or something…it is like this…15y^2+19y+6…I see that as being prime but I am curious as to how you factor a problem like that because it was the only one like that on the final and I wonder if it is something you have to know for Intermediate Algebra…the pattern to that problem looks like it is Intermediate Algebra level or something. So, I am just curious.

OK, trying to factor 15y^2 + 19y + 6

(A) Yes, Sue is right, ALWAYS look for a common factor first.
Since 19 is a prime number (nothing divides it evenly except itself and 1) there is no common number, and since the last term is just 6 without a letter there is no common letter.
So this one has no common factors.

(b) Count the terms.
Two terms — *common factor* OR *difference of squares* or cubes, ONLY possibilities
Three terms — *binomial pattern*
Four terms — try *grouping* two and two with common factors in each pair; very very rarely a trick three and one, don’t worry about it at this point.
This one is three terms, look at binomial pattern (__y + __)(__y + __)

(c) Fill in the blanks.
Look at the first (y^2) term. How many ways can we make 15?
15 = 1 x 15, 15 = 3 x 5, 15 = 5 x 3, 15 = 15 x 1
(note: to get *all* possible factorings of a number, just work up through it — 15 = 1 X 15, 15 = 2 x nope, doesn’t work, 15 = 3 x 5, 15 = 4 x nope doesn’t work, 15 = 5 x 3 hey I’m starting to repeat myself, got all the factorings, don’t forget to write 15 = 15 x 1.)
Since 3 x 5 is the same as 5 x 2, we really only have two choices. But when doing this, never forget the 1 x pattern! Rare but it happens.
Look at the last (constant number) term. How many ways can we make 6?
6 = 1 x 6, 6 = 2 x 3, 6 = 3 x 2, 6 = 6 x 1
Now we’s better keep all the possible orders because when we combinte with the factors of 15 we get different pairings, (3y + 2)(5y + 3) gives a different result from (3y + 3)(5y + 2) – multiply out those two and check them; we know the first and last terms will be right because we built them that way, but the middle terms will be different.

(d) Two possible first pairs times four possible last pairs gives eight combinations to test. Not too bad, but we can reduce things even further.
**We don’t ever need to bother with pairs like (3y + 3) with a common factor — IF we removed all common factors to begin with, then we wouldn’t have a common factor in the factoring result. This leaves four combos to test, not bad at all.
** No guarantees, but this is the way to bet: Start with the middle values, 3 x 5 = 15 and 2 x 3 = 6. *Only* if these don’t work go to the extreme values like 15 x 1 = 15 and 1 x 6 = 6. Work out from midddling values towards more extremes. MOST of the time school problems (and real problems if they factor at all) fall in the less extreme group.

(e) List pairs to test, in order from most likely middling value to extreme values, and crossing out any common factors that come up in the pairing
(3y + 2)(5y + 3)
(Omit (3y + 3)(5y + 2), common factor)
(3y + 1)(5y + 6)
(Omit (3y + 6)(5y + 1), common factor)
(Omit (y + 2)(15y + 3), common factor)
(y + 3)(15y + 2)
(Omit (y + 1)(15y + 6), common factor)
(y + 6)(15y + 1)

(f) Multiply, in turn, in order — with practice you can do this in your head — and look at that middle term (We know first and last are right because we built them that way, so look at the middle)

First trial factoring
(3y + 2)(5y + 3) = 15y^2 + 9y + 10y + 6 = 15y^2 + 19y + 6
There’s that 19y !! Got it in one, first try.

ANSWER: 15y^2 + 19y + 6 = (3y + 2)(5y + 3)

If you use this decision-making process, which is logical and sensible and once you follow it quite organized, you can get *most* factorings in two or three tries on average. And you can do many of them in your head, amazing your friends and neighbours. This is much much quicker than the trick methods sometime taught.

The other nice thing about working through a logical process is that you can *prove* without a shadow of a doubt when a trinomial is prime or unfactorable. As long as you get all common factors first and get every possible factoring of the numbers and try all possible pairings (which as above is not too too many), you know for sure that you have tried everything possible and there is no trick factoring left.

Do you need factoring for Intermediate Algebra? Yes.
Math is a cumulative subject. Each level depends on all the stairsteps below leading up to it. Cut out stair #3 and all the stairs from 4 up come crashing down on you.
In Intermediate Algebra, problems will be solved by factoring, and the professor will not stop to explain it, or will at most give you the thirty-second summary. Don’t blame the professor; he has a grand total of 42 hours (work it out, how many class hours you actually get) to cover that whole huge textbook, and he is working flat-out to even come close. Also he has more than enough of this term’s work to teach, he doesn’t have time to reteach last term and the one before it as well. So yes, factoring was taught this term because it’s going to be used next term; actually kind of nice and reassuring that work is taught for a reason and not going to waste.

15y^2 + 19y + 6

Here’s a “grouping” method that always works and is systematic. Multiply the first and last numbers:

Fifteen x 6 = 90.

Make a chart of the possible factors of 90. Your goal is to find two of them that add up to 19.

(For *any* trinomial, you’d wnat a pair of numbers that has the product equal to the product of the first and last number, and a sum equal to the middle number.)

1 90
2 45
3 30
5 18
6 15
9 10

…. Bingo…..

Now, rewrite the problem with the middle term split up. INstead of “19x” write “10x + 9x” because it’s the same thing.

15x^2 + 10x + 9x + 6

Now, factor by grouping.

5x (3x+2) + 3(3x + 2)

(3x + 2) is one of your factors: 5x + 3 is the other.

(5x + 3) (3x + 2)

I find this is a godsend when the ol’ anxiety kicks in and you want STeps That Really Always Work (STRAW :-)) especially for linear thinkers for whom “trying all the possible combinations” seems messy.

Sue — I frequently get students coming to me with a “really great fast way” to solve problems that I have been trying to teach them. OK, I say, show it to me. Well, they say, you take this number here and shove it there — no, wait, you take this number and multiply it by that one and add this here — no, I don’t know, but it’s a really great method, so much faster than what you showed me. My answer is yeah, but you have a half a chance to remember my method because it goes step by step, while you have just proved to me that this so-called “fast” method is *really* infinitely slow because you can’t ever get it right.
Your factoring thingy above, the slope-point formula, FOIL, picking up numbers and moving them in equations and sometimes changing the sign and sometimes not, various forms of “cancellation”, and my least favourite, “cross-multiplication”, all come in this category. I disrecommend them strongly for the first-year student because quite simply, they don’t work. Not for the beginner. The first-year student doesn’t know in what cases the trick works and when it doesn’t, and also has a memory overload of too many unrelated tricks with no hook to hang them on (is this the one where you multipy things together or the one where you square everything or …) so you get memory crashes on tests. Worse, you get students arriving in Algebra 2 with a mind pretty much blank except for a vague memory that you square something somewhere.

Perhaps because of the particular sequence here (everybody has had a fair amount of practice with trinomials where the first coefficient is 1, and they have had practice factoring by grouping because the book has a bunch of problems already set up that way), the “sum and product” method has worked really well because it’s so much less demanding on the working memory than the “try all the possible combiinations for the first coefficient, and then all the possible combinations for the third coefficient, and then see which ones work.”

I’ve seen other strategies have exactly the same confounding effect you describe — “There’s a trick… let’s see…” — and my most frustrating is the textbook ADHD person who will impulsively head off into a randomly selected strategy having nothing to do with the actual problem (probably one of the clearest manifestations of the executive function issues — the whole ‘reasoning’ process just doesnt’ kick in, even though it’s there and she doesn’t *want* to waste all this time…)

Can y ou open a .doc file?

I scanned the puppy in at http://www.resourceroom.net/math/factoringflowchart.doc

Let me know if it’s gotta be a .jpg or something — .doc was the easiest to manage.